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ELEMENTS 


or 


PLANE  AND  SOLID  GEOMETRY. 


BY 


G.  A.  WENT  WORTH,  A.  M., 

PROFESSOR  OF  MATHEMATICS   IN   PHILLIPS  EXETER  ACADEMY. 


BOSTON: 
PUBLISHED  BY   GINN,   HEATH,   &  CO. 

18.85.    . 


*-<LdX< 


Copyright,  1877. 
By   GINN    AND    HEATH. 


j.  s.  cushing, 

Superintendent  of  Printino, 
ioi  Pearl  St.,  Boston. 


PREFACE 


Most  persons  do  not  possess,  and  do  not  easily  acquire,  the 
power  of  abstraction  requisite  for  apprehending  the  Geometri- 
cal conceptions,  and  for  keeping  in  mind  the  successive  steps 
of  a  continuous  argument.  Hence,  with  a  very  large  proportion 
of  beginners  in  Geometry,  it  depends  mainly  upon  the  form  in 
which  the  subject  is  presented  whether  they  pursue  the  study 
with  indifference,  not  to  say  aversion,  or  with  increasing  interest 
and  pleasure. 

In  compiling  the  present  treatise,  this  fact  has  been  kept  con- 
stantly in  view.  All  unnecessary  discussions  and  scholia  hav« 
been  avoided ;  and  such  methods  have  been  adopted  as  experi- 
ence and  attentive  observation,  combined  with  repeated  trials, 
have  shown  to  be  most  readily  comprehended.  No  attempt  has 
been  made  to  render  more  intelligible  the  simple  notions  of 
position,  magnitude,  and  direction,  which  every  child  derives 
from  observation  ;  but  it  is  believed  that  these  notions  have 
been  limited  and  defined  with  mathematical  precision. 

A  few  symbols,  which  stand  for  Avords  and  not  fur  operations, 
have  been  used,  but  these  are  of  so  great  utility  in  giving  style 
and  perspicuity  to  the  demonstrations  that  no  apology  seems 
necessary  for  their  introduction. 

Great  pains  have  been  taken  to  make  the  page  attractive. 
The  figures  are  large  and  distinct,  and  are  placed  in  the  middle 
of  the  page,  so  that  they  fall  directly  under  the  eye  in  imme- 
diate connection  with  the  corresponding  text.     The  given  lines 


M306198 


IV  PREFACE. 


of  the  figures  are  full  lines,  the  lines  employed  as  aids  in  the 
demonstrations  are  short-dotted,  and  the  resulting  lines  are  long- 
dotted. 

In  each  proposition  a  concise  statement  of  what  is  given  is 
printed  in  one  kind  of  type,  of  what  is  required  in  another,  and 
the  demonstration  in  still  another.  The  reason  for  each  step 
is  indicated  in  small  type  between  that  step  and  the  one  follow- 
ing, thus  preventing  the  necessity  of  interrupting  the  process  of 
the  argument  by  referring  to  a  previous  section.  The  number 
of  the  section,  however,  on  which  the  reason  depends  is  placed 
at  the  side  of  the  page.  The  constituent  parts  of  the  propo- 
sitions are  carefully  marked.  Moreover,  each  distinct  assertion  in 
the  demonstrations,  and  each  particular  direction  in  the  construc- 
tions of  the  figures,  begins  a  new  line ;  and  in  no  case  is  it  neces- 
sary to  turn  the  page  in  reading  a  demonstration. 

This  arrangement  presents  obvious  advantages.  The  pupil 
perceives  at  once  what  is  given  and  what  is  required,  readily 
refers  to  the  figure  at  every  step,  becomes  perfectly  familiar  with 
the  language  of  Geometry,  acquires  facility  in  simple  and  accu- 
rate expression,  rapidly  learns  to  reason,  and  lays  a  foundation 
for  the  complete  establishing  of  the  science. 

A  few  propositions  have  been  given  that  might  properly  bo 
considered  as  corollaries.  The  reason  for  this  is  the  great  diffi- 
culty of  convincing  the  average  student  that  any  importance 
should  be  attached  to  a  corollary.  Original  exercises,  however, 
have  been  given,  not  too  numerous  or  too  difficult  to  discourage 
the  beginner,  but  well  adapted  to  afford  an  effectual  test  of  the 
degree  in  which  he  is  mastering  the  subjects  of  his  reading. 
Some  of  these  exercises  have  been  placed  in  the  early  part  of 
the  work  in  order  that  the  student  may  discover,  at  the  outset, 
that  to  commit  to  memory  a  number  of  theorems  and  to  repro- 
duce them  in  an  examination  is  a  useless  and  pernicious  labor ; 
but  to  learn  their  uses  and  applications,  and  to  acquire  a  readi- 
ness in  exemplifying  their  utility,  is  to  derive  the  full  benefit 
of  that  mathematical  training  which  looks  not  so  much  to  the 


PREFACE. 


attainment  of  information  as  to  the  discipline  of  the  mental  fac- 
ulties. 

It  only  remains  to  express  my  sense  of  obligation  to  Dr. 
D.  F.  Wells  for  valuable  assistance,  and  to  the  University 
Press  for  the  elegance  with  which  the  book  has  been  printed  ; 
and  also  to  give  assurance  that  any  suggestions  relating  to  the 
work  will  be  thankfully  received. 

G.  A.  WENTWORTH. 

Phillips  Exeter  Academy, 
January,  1878. 


NOTE  TO   THIRD   EDITION. 

In  this  edition  I  have  endeavored  to  present  a  more  rigor- 
ous, but  not  less  simple,  treatment  of  Parallels,  Ratio,  and 
Limits.  The  changes  are  not  sufficient  to  prevent  the  simulta 
neous  use  of  the  old  and  new  editions  in  the  class ;  still  they  arc 
very  important,  and  have  been  made  after  the  most  careful  and 
prolonged  consideration. 

I  have  to  express  my  thanks  for  valuable  suggestions  received 
from  many  correspondents  ;  and  a  special  acknowledgment  is  due 
from  me  to  Professor  C.  H.  Judson,  of  Furman  University, 
Greenville,  South  Carolina,  to  whom  I  am  indebted  for  assist- 
ance in  effecting  many  improvements  in  this  edition. 

TO  THE  TEACHER. 

When  the  pupil  is  reading  each  Book  for  the  first  time,  it  will  be 
well  to  let  him  write  his  proofs  on  the  blackboard  in  his  own  lan- 
guage ;  care  being  taken  that  his  language  be  the  simplest  possible, 
that  the  arrangement  of  work  be  vertical  (without  side  work),  and 
that  the  figures  be  accurately  constructed. 

This  method  will  furnish  a  valuable  exercise  as  a  language  lesson, 
will  cultivate  the  habit  of  neat  and  orderly  arrangement  of  work, 
and  will  allow  a  brief  interval  for  deliberating  on  each  step. 

After  a  Book  has  been  read  in  this  way  the  pupil  should  review 
the  Book,  and  should  be  required  to  draw  the  figures  free-hand.     He 


Vi  PREFACE. 


should  state  and  prove  the  propositions  orally,  using  a  pointer  to 
indicate  on  the  figure  every  line  and  angle  named.  He  should  be 
encouraged,  in  reviewing  each  Book,  to  do  the  original  exercises  ;  to 
state  the  converse  of  propositions  ;  to  determine  from  the  statement, 
if  possible,  whether  the  converse  be  true  or  false,  and  if  the  converse 
be  true  to  demonstrate  it ;  and  also  to  give  well-considered  answers 
to  questions  which  may  be  asked  him  on  many  propositions. 

The  Teacher  is  strongly  advised  to  illustrate,  geometrically  and 
arithmetically,  the  principles  of  limits.  Thus  a  rectangle  with  a 
constant  base  6,  and  a  variable  altitude  x,  will  afford  an  obvious 
illustration  of  the  axiomatic  truth  contained  in  [4],  page  88.  If  x 
increase  and  approach  the  altitude  a  as  a  limit,  the  area  of  the  rec- 
tangle increases  and  approaches  the  area  of  the  rectangle  a  b  as  a 
limit ;  if,  however,  x  decrease  and  approach  zero  as  a  limit,  the  area 
of  the  rectangle  decreases  and  approaches  zero  for  a  limit.  An  arith- 
metical illustration  of  this  truth  would  be  given  by  multiplying  a 
constant  into  the  approximate  values  of  any  repetend.  If,  for  exam- 
ple, we  take  the  constant  60  and  the  repetend  .3333,  etc.,  the  approxi- 
mate values  of  the  repetend  will  be  fa  ffa  iWr»  l^A%  etc>  anc^ 
these  values  multiplied  by  60  give  the  series  18,  19.8,  19.98,  19.998, 
etc.,  which  evidently  approach  20  as  a  limit ;  but  the  product  of  60 
into  £  (the  limit  of  the  repetend  .333,  etc.)  is  also  20. 

Again,  if  we  multiply  60  into  the  different  values  of  the  decreasing 
series,  fa  -gfa,  sfaj,  3iri5inr>  e*c-i  which  approaches  zero  as  a  limit, 
we  shall  get  the  decreasing  series,  2,  \,  fa,  jfo,  etc.  ;  and  this  series 
evidently  approaches  zero  as  a  limit. 

In  this  way  the  pupil  may  easily  be  led  to  a  complete  comprehen- 
sion of  the  whole  subject  of  limits. 

The  Teacher  is  likewise  advised  to  give  frequent  written  examina- 
tions. These  should  not  be  too  difficult,  and  sufficient  time  should 
be  allowed  for  accurately  constructing  the  figures,  for  choosing  the 
best  language,  and  for  determining  the  best  arrangement. 

The  time  necessary  for  the  reading  of  examination-books  will  be 
diminished  by  more  than  one-half,  if  the  use  of  the  symbols  employed 
in  this  book  be  permitted. 

G.  A.  W. 

Phillips  Exeter  Academy, 
January,  1879. 


CONTENTS. 


PLANE   GEOMETEY. 


BOOK  I.    Rectilinear  Figures. 

Pagr 

Introductory  Remarks  

.        3 

Definitions 

4 

Straight  Lines         

6 

Plane  Angles 

7 

Angular  Magnitude 

9 

Superposition 

10 

Mathematical  Terms 

.     11 

Axioms  and  Postulates 

12 

Symbols  and  Abbreviations    .... 

.     13 

Perpendicular  and  Oblique  Links    . 

14 

Parallel  Lines 

.     24 

Triangles 

37 

Quadrilaterals 

.     58 

Polygons  in  General 

63 

BOOK  II.    Circles. 

Definitions 

73 

Straight  Lines  and  Circles  .... 

.     75 

Measurement 

86 

Theory  of  Limits 

.     87 

Supplementary  Propositions      .... 

100 

Constructions 

.   103 

BOOK  III.    Proportional  Lines  and  Similar  Polygons. 

Theory  of  Proportion .128 

Proportional  Lines 139 

Similar  Polygons 143 

Constructions 164 


CONTENTS. 


BOOK  IV.    Comparison  and  Measurement  of  the  Sur- 
faces of  Polygons. 

Comparison  and  Measurement  of  Polygons       .        .        .174 

Constructions 194 

BOOK  V.    Regular  Polygons  and  Circles. 

Regular  Polygons  and  Circles 210 

Constructions 224 

Isoperimetrical  Polygons.     Supplementary       .        .        .  237 

Symmetry.     Supplementary 245 


SOLID    GEOMETRY. 

BOOK  VI.    Planes  and  Solid  Angles. 

Lines  and  Planes 251 

Dihedral  Angles ,  268 

Supplementary  Propositions 275 

Polyhedral  Angles 277 

BOOK  VII.    Polyhedrons,  Cylinders,  and  Cones. 

Prisms 286 

Pyramids 302 

Similar  Polyhedrons 317 

PvEGUlar  Polyhedrons        .        .        .        .        .        .        .  322 

Supplementary  Propositions 326 

Cylinders 328 

Cones 339 

BOOK  VIII.    The  Sphere. 

Sections  and  Tangents 349 

Distances  on  the  Surface  of  the  Sphere     .        .        .  356 

Spherical  Angles  .        .        .        .        .        .        .        .     .   .  363 

Spherical  Polygons  and  Pyramids 365 

Comparison  and  Measurement  of  Spherical  Surfaces     .  383 

Volume  or  the  Sphere 396 


ELEMENTS  OF  GEOMETRY. 


BOOK  I. 

RECTILINEAR   FIGURES. 


Introductory  Remarks. 

A  rough  block  of  marble,  under  the  stone-cutter's  hammer, 
may  be  made  to  assume  regularity  of  form. 

If  a  block  be  cut  in  the  shape   repre-  ^ y 

sented  in  this  diagram, 

It  will  have  six  flat  faces. 

Each  face  of  the  block  is  called  a  Sur- 
face. 

If  these  surfaces  be  made  smooth  by  pol- 
ishing, so  that,  when  a  straight-edge  is  applied  to  any  one  of 
them,  the  straight-edge  in  every  part  will  touch  the  surface,  the 
surfaces  are  called  Plane  Surfaces. 

The  sharp  edge  in  which  any  two  of  these  surfaces  meet  is 
called  a  Line. 

The  place  at  which  any  three  of  these  lines  meet  is  called  a 
Point. 

If  now  the  block  be  removed,  we  may  think  of  the  place 
occupied  by  the  block  as  being  of  precisely  the  same  shape  and 
size  as  the  block  itself;  also,  as  having  surfaces  or  boundaries 
which  separate  it  from  surrounding  space.  We  may  likewise 
think  of  these  surfaces  as  having  lines  for  their  boundaries  or 
limits ;  and  of  these  lines  as  having  points  for  their  extremities 
or  limits. 

A  Solid,  as  the  term  is  used  in  Geometry,  is  a  limited  por- 
tion of  space. 

After  we  acquire  a  clear  notion  of  surfaces  as  boundaries  of 
solids,  we  can  easily  conceive  of  surfaces  apart  from  solids,  and 


GEOMETRY. BOOK   I. 


suppose  them  of  unlimited  extent.  Likewise  we  can  conceive  of 
lines  apart  from  surfaces,  and  suppose  them  of  unlimited  length; 
of  points  apart  from  lines  as  having  position,  but  no  extent. 

Definitions. 

1.  Def.  Space  or  Extension  has  three  Dimensions,  called 
Length,  Breadth,  and  Thickness. 

2.  Def.    A  Point  has  position  without  extension. 

3.  Def.  A  Line  has  only  one  of  the  dimensions  of  exten- 
sion, namely,  length. 

The  lines  which  we  draw  are  only  imperfect  representations 
of  the  true  lines  of  Geometry. 

A  line  may  be  conceived  as  traced  or  generated  by  a  point  in 
motion. 

4.  Def.  A  Surface  has  only  two  of  the  dimensions  of  ex- 
tension, length  and  breadth. 

A  surface  may  be  conceived  as  generated  by  a  line  in  motion. 

5.  Def.  A  Solid  has  the  three  dimensions  of  extension, 
length,  breadth,  and  thickness.  Hence  a  solid  extends  in  all  direc- 
tions. 

A  solid  may  be  conceived  as  generated  by  a  surface  in  motion. 

Thus,  in  the  diagram,  let  the  upright 
surface  A  B  CD  move  to  the  right  to 
the  position  E  F  H  K.  The  points 
A,  B,  C,  and  D  will  generate  the  lines 
AE,  BF,  CK,  and  D  H  respectively.         C  * 

And  the  lines  A  B,  B  D,  DC,  and  A  C  will  generate  the  sur- 
faces A  F,  B  H,  D  K,  and  A  K  respectively.  And  the  surface 
ABC D  will  generate  the  solid  A  H. 

The  relative  situation  of  the  two  points  A  and  H  involves 
three,  and  only  three,  independent  elements.  To  pass  from  A  to  H 
it  is  necessary  to  move  East  (if  we  suppose  the  direction  A  E  to 


!       I 


]H 


DEFINITIONS. 


be  due  East)  a  distance  equal  to  A  E,  North  a  distance  equal  to 
E  F,  and  down  a  distance  equal  to  F  H. 

These  three  dimensions  we  designate  for  convenience  length, 
breadth,  and  thickness. 

6.  The  limits  (extremities)  of  lines  are  points. 
The  limits  (boundaries)  of  surfaces  are  lines. 
The  limits  (boundaries)  of  solids  are  surfaces. 

7.  Def.    Extension  is  also  called  Magnitude. 

When  reference  is  had  to  extent,  lines,  surfaces,  and  solids  are 
called  magnitudes. 

8.  Def.    A  Straight  line  is  a  line  which  has 
the  same  direction  throughout  its  whole  extent. 

9.  Def.  A  Curved  line  is  a  line  which  changes 
its  direction  at  every  point. 

10.  Def.    A  Broken  line  is  a  series  of  con- 
nected straight  lines. 

When  the  word  line  is  used  a  straight  line  is  meant;  and 
when  the  word  curve  is  used  a  curved  line  is  meant. 

1 1.  Def.  A  Plane  Surface,  or  a  Plane,  is  a  surface  in  which, 
if  any  two  points  be  taken,  the  straight  line  joining  these  points 
will  lie  wholly  in  the  surface. 

12.  Def.  A  Curved  Surface  is  a  surface  no  part  of  which 
is  plane. 

1 3.  Figure  or  form  depends  upon  the  relative  position  of 
points.  Thus,  the  figure  or  form  of  a  line  (straight  or  curved) 
depends  upon  the  relative  position  of  points  in  that  line ;  the 
figure  or  form  of  a  surface  depends  upon  the  relative  position  of 
points  in  that  surface. 

When  reference  is  had  to  form  or  shape,  lines,  surfaces,  and 
solids  are  called  figures. 


6  GEOMETRY. BOOK   I. 

1 4.  Def.  A  Plane  Figure  is  a  figure,  all  points  of  which 
are  in  the  same  plane. 

15.  Def.  Geometry  is  the  science  which  treats  of  position, 
magnitude,  and  form. 

Points,  lines,  surfaces,  and  solids,  with  their  relations,  arc 
the  geometrical  conceptions,  and  constitute  the  subject-matter  of 
Geometry. 

16.  Plane  Geometry  treats  of  plane  figures. 

Plane  figures  are  either  rectilinear,  curvilinear,  or  mixtilinear. 

Plane  figures  formed  by  straight  lines  are  called  rectilinear 
figures ;  those  formed  by  curved  lines  are  called  curvilinear  fig- 
ures ;  and  those  formed  by  straight  and  curved  lines  are  called 
mixtilinear  figures. 

17.  Def.  Figures  which  have  the  same  form  are  called 
Similar  Figures.  Figures  which  have  the  same  extent  are  called 
Equivalent  Figures.  Figures  which  have  the  same  form  and 
extent  are  called  Equal  Figures. 


On  Straight  Lines. 

18.  If  the  direction  of  a  straight  line  and  a  point  in  the 
line  be  known,  the  position  of  the  line  is  known;  that  is,  a 
straight  line  is  determined  in  position  if  its  direction  and  one  of 
its  points  be  known. 

Hence,  all  straight  lines  which  pass  through  the  same  point  in 
the  same  direction  coincide. 

Between  two  points  one,  and  but  one,  straight  line  can  be 
drawn ;  that  is,  a  straight  line  is  determined  in  position  if  two  of 
its  points  be  known. 

Of  all  lines  between  two  points,  the  shortest  is  the  straight 
iine ;  and  the  straight  line  is  called  the  distance  between  the 
two  points. 


DEFINITIONS. 


The  point  from  which  a  line  is  drawn  is  called  its  origin. 

19.  If  a  line,  as  G  B,  ± % £,  be  produced  through  C, 

the  portions  GB  and  G  A  may  be  regarded  as  different  lines 
having  opposite  directions  from  the  point  G. 

Hence,  every  straight  line,  as  A  B,  £ £ ,  has  two  opposite 

directions,  namely  from  A  toward  B,  which  is  expressed  by  say- 
ing line  A  B,  and  from  B  toward  A,  which  is  expressed  by 
saying  line  B  A. 

20.  If  a  straight  line  change  its  magnitude,  it  must  become 

longer  or  shorter.     Thus  by  prolonging  A  B  to  C,  A      f ?, 

AC  =  AB  +  BC;  and  conversely,  BG  =  AG~AB. 

If  a  line  increase  so  that  it  is  prolonged  by  its  own  magnitude 
several  times  in  succession,  the  line  is  multiplied,  and  the  result- 
ing line  is  called  a  multiple  of  the  given  line.     Thus,  if  A  B  = 

BG  =  G  D,  etc.,  £_* % £_!,   then  A  C=  2  A  B,  AB  = 

3AB,  etc. 

It  must  also  be  possible  to  divide  a  given  straight  line  into  an 
assigned  number  of  equal  parts.  For,  assumed  that  the  »th 
part  of  a  given  line  were  not  attainable,  then  the  double,  triple, 
quadruple,  of  the  nth  part  would  not  be  attainable.  Among 
these  multiples,  however,  we  should  reach  the  nth  multiple  of 
this  nth  part,  that  is,  the  line  itself.  Hence,  the  line  itself  would 
not  be  attainable  ;  which  contradicts  the  hypothesis  that  we  have 
the  given  line  before  us. 

Therefore,  it  is  always  possible  to  add,  subtract,  multiply,  and 
divide  lines  of  given  length. 

21.  Since  every  straight  line  has  the  property  of  direction, 
it  must  be  true  that  two  straight  lines  have  either  the  same 
direction  or  different  directions. 

Two  straight  lines  which  have  the  same  direction,  without  coin- 
ciding, can  never  meet ;  for  if  they  could  meet,  then  we  should 
have  two  straight  lines  passing  through  the  same  point  in  the 
same  direction.     Such  lines,  however,  coincide.  §  18 


8  GEOMETRY. BOOK    I. 

22.  Two  straight  lines  which  lie  in  the  same  plane  and  have 
different  directions  must  meet  if  sufficiently  prolonged  ;  and  must 
have  one,  and  but  one,  point  in  common. 

Conversely  :  Two  straight  lines  lying  in  the  same  plane  ivhich 
do  not  meet  have  the  same  direction;  for  if  they  had  different 
directions  they  would  meet,  which  is  contrary  to  the  hypothesis 
that  they  do  not  meet. 

Two  straight  lines  which  meet  have  different  directions;  for 
if  they  had  the  same  direction  they  would  never  meet  (§  21), 
which  is  contrary  to  the  hypothesis  that  they  do  meet. 


On  Plane  Angles. 

23.  Def.  An  Angle  is  the  difference  in  direction  of  two 
lines.  The  point  in  which  the  lines  (prolonged  if  necessary) 
meet  is  called  the  Vertex,  and  the  lines  are  called  the  Sides  of 
the  angle. 

An  angle  is  designated  by  placing  a  letter  at  its  vertex,  and 
one  at  each  of  its  sides.  In  reading,  we  name  the  three  let- 
ters, putting  the  letter  at  the  vertex  between  the  other  two.  When 
the  point  is  the  vertex  of  but  one  angle  we  usually  name  the 
letter  at  the  vertex  only ;  thus,  in  Fig.  1,  we  read  the  angle  by 


calling  it  angle  A.  But  in  Fig.  2,  H  is  the  common  vertex  of 
two  angles,  so  that  if  we  were  to  say  the  angle  H,  it  would  not 
be  known  whether  we  meant  the  angle  marked  3  or  that 
marked  4.  We  avoid  all  ambiguity  by  reading  the  former  as 
the  angle  E  H  D,  and  the  latter  as  the  angle  E  H  F. 


DEFINITIONS. 


9 


angles 


D 


The  magnitude  of  an  angle  depends  wholly  upon  the  extent 
of  opening  of  its  sides,  and  not  upon  their 
length.  Thus  if  the  sides  of  the  angle  B  AC, 
namely,  A  B  and  A  C,  be  prolonged,  their 
extent  of  opening  will  not  be  altered,  and  the 
size  of  the  angle,  consequently,  will  not  be 
changed. 

24.  Def.    Adjacent  Angles  are  angles 
having  a  common  vertex  and  a  common 
side   between    them.      Thus    the 
C D E  and  CDF  are  adjacent  angles. 

25.  Def.  A  Right  Angle  is  an  angle  included  between  two 
straight  lines  which  meet  each  other  so  that  the  two  adjacent 
angles  formed  by  producing  one  of  the  lines 

through  the  vertex  are  equal.  Thus  if  the 
straight  line  A  B  meet  the  straight  line  C  D 
so  that  the  adjacent  angles  A BC  and  ABD 
are  equal  to  one  another,  each  of  these  an- 
gles is  called  a  right  angle. 

26.  Def.  Perp>endicular  Lines  are  lines 
which  make  a  right  angle  with  each  other. 

27.  Def.  An  Acute  Angle  is  an  angle 
less  than  a  right  angle ;  as  the  angle  B  A  C. 

28.  Def.  An  Obtuse  Angle  is  an  angle 
greater  than  a  right  angle ;  as  the  angle 
DEF. 

29.  Def.    Acute  and  obtuse  angles,  in 
distinction  from  right  angles,  are  called  ob- 
lique angles  ;  and  intersecting  lines  which  are  not  perpendicular 
to  each  other  are  called  oblique  lines. 

30.  Def.  The  Complement  of  an  angle  is 
the  difference  between  a  right  angle  and  the 
given  angle.     Thus  A  B  D  is  the  complement 


B 


D 


D 


of  the  angle  BBC;  also  D  B  C  is,  the  com- 
plement of  the  angle  ABD. 


10 


GEOMETRY. BOOK   I. 


31.  Def.  The  Supplement  of  an  angle 
is  the  difference  between  two  right  angles 
and  the  given  angle.  Thus  A  CD  is  the 
supplement  of  the  angle  DC B\  also  D  C B 
is  the  supplement  of  the  angle  AC  D. 

32.  Def.  Vertical  Angles  are  angles 
which  have  the  same  vertex,  and  their 
sides  extending  in  opposite  directions. 
Thus  the  angles  A  OB  and  COB  are 
vertical  angles,  as  also  the  angles  A  0  C 
sm&DOB. 


1> 


B 


On  Angular  Magnitude. 


,1' 


0 


B> 


C 
A 


33.  Let  the  lines  B  B'  and  A  A'  be  in  b 

the  same  plane,  and  let  B  B'  be  perpen- 
dicular to  A  A'  at  the  point  0. 

Suppose  the  straight  line  0  C  to  move 
in  this  plane  from  coincidence  with  0  A, 
about  the  point  0  as  a  pivot,  to  the  po- 
sition 0  C ;  then  the  line  0  C  describes  or 
generates  the  angle  A  0  C. 

The  amount  of  rotation  of  the  line,  from  the  position  0  A  to 
the  position  0  C,  is  the  Angular  Magnitude  A  0  C. 

If  the  rotating  line  move  from  the  position  0  A  to  the  po- 
sition 0  B,  perpendicular  to  0  A,  it  generates  a  right  angle  ;  to 
the  position  0  A'  it  generates  two  right  angles ;  to  the  position 
OB',  as  indicated  by  the  dotted  line,  it  generates  three  right 
angles;  and  if  it  continue  its  rotation  to  the  position  0 A, 
whence  it  started,  it  generates  four  right  angles. 

Hence  the  whole  angular  magnitude  about  a  point  in  a  plane 
is  equal  to  four  right  angles,  and  the  angular  magnitude  about 
a  point  on  one  side  of  a  straight  line  drawn  through  that  point 
is  equal  to  two  right  angles. 


DEFINITIONS. 


11 


0 


// 


o 

Fig.  2. 


34.  Now  since  the  augular  magnitude  about  the  point  0  is 
neither  increased  nor  diminished  by  the  number  of  lines  which 
radiate  from  that  point,  the  sum  of  all  the  angles  about  a  point 
in  a  plane,  as  AOB+BOC+COD,  etc.,  in  Fig.  1,  is  equal 
to  four  right  angles  ;  and  the  sum  of  all  the  angles  about  a  point 
on  one  side  of  a  straight  line  drawn  through  that  point,  as 
AOB+BOC+COD,  etc.,  Fig.  2,  is  equal  to  two  right 
angles. 

Hence  two  adjacent  angles,  OCA  and  OGB,  jy 

formed  by  two  straight  lines,  of  which  one  is 
produced  from  the  point  of  meeting  in  both  di- 
rections, are  supplements  of  eacli  other,  and  may  J 
be  called  supplementary  adjacent  angles. 

On  the  Method  of  Superposition. 

35.  The  test  of  the  equality  of  two  geometrical  magnitudes 
is  that  they  coincide  point  for  point. 

Thus,  two  straight  lines  are  equal,  if  they  can  be  so  placed 
that  the  points  at  their  extremities  coincide.  Two  angles  arc 
equal,  if  they  can  be  so  placed  that  their  vertices  coincide  in 
position  and  their  sides  in  direction. 

In  applying  this  test  of  equality,  we  assume  that  a  line  may 
be  moved  from  one  place  to  another  without  altering  its  length ; 
that  an  angle  may  be  taken  up,  turned  over,  and  put  down, 
without  altering  the  difference  in  direction  of  its  sides. 


12 


GEOMETRY. BOOK    I. 


This  method  enables  us  to  com- 
pare unequal  magnitudes  of  the 
same  kind.  Suppose  we  have  two 
angles,  ABC  and  A'  B'  C.  Let 
the  side  B  C  be  placed  on  the  side 
B'  C,  so  that  the  vertex  B  shall  fall  on  B',  then  if  the  side  B  A 
fall  on  B'  A1,  the  angle  ABC  equals  the  angle  A'  B'  C ;  if  the 
side  B  A  fall  between  B'  C  and  B'  A'  in  the  direction  B'  D,  the 
angle  A  B  C  is  less  than  A'  B'  C"  ;  but  if  the  side  B  A  fall  in  the 
direction  B'E,  the  angle  A  B  C  is  greater  than  A'  B'  C. 

This  method  of  superposition  en-  B  q 

ables  us  to  add  magnitudes  of  the 

same  kind.     Thus,  if  we  have  two     c D 

straight   lines    A  B  and    CD,    by     A B 

placing  the  point  C  on  B,  and  keeping  C  D  in  the  same  direc- 
tion with  A  B,  we  shall  have  one  continuous  straight  line  A  D 
equal  to  the  sura  of  the  lines  A  B 
and  C  D. 

Again  :  if  we  have  the  angles 
ABC  and  D  E  F,  by  placing 
the  vertex  B  on  E  and  the  side 
BC  in  the  direction  of  ED,  the 
angle  ABC  will  take  the  position 
A  ED,  and  the  angles  D E F  and 
ABC  will  together  equal  the  an- 
gle AEF. 


Mathematical  Terms. 

36.  Def.  A  Demonstration  is  a  course  of  reasoning  by  which 
the  truth  or  falsity  of  a  particular  statement  is  logically  established. 

37.  Def.    A  Theorem  is  a  truth  to  be  demonstrated. 

38.  Def.    A  Construction  is   a  graphical  representation  of 
a  geometrical  conception. 

39.  Def.    A  Problem  is  a  construction  to  be  effected,  or  a 
question  to  be  investigated. 


DEFIXITIONS.  13 


40.  Def.    An  Axiom  is  a  truth  which  is  admitted  without 
demonstration. 

41.  Def.   A  Postulate  is  a  problem  which  is  admitted  to 
be  possible. 

42.  Def.    A  Proposition  is  either  a  theorem  or  a  problem. 

43.  Def.    A  Corollary  is  a  truth  easily  deduced  from  the 
proposition  to  which  it  is  attached. 

44.  Def.    A  Scholium  is  a  remark  upon  some  particular  fea- 
ture of  a  proposition. 

45.  Def.    An  Hypothesis   is   a   supposition   made   in   the 
enunciation  of  a  proposition,  or  in  the  course  of  a  demonstration. 

46.  Axioms. 

1.  Tilings  which  are  equal  to  the  same  thing  are  equal  to  each 

other. 

2.  When  equals  are  added  to  equals  the  sums  are  equal. 

3.  When  equals  are  taken  from  equals  the  remainders  are  equal. 

4.  When  equals  are  added  to  unequals  the  sums  are  unequal. 

5.  When  equals  are  taken  from  unequals  the  remainders  are 

unequal. 

6.  Things  which  are  double  the  same  tiling,  or  equal  things, 

are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,   or  of  equal 

things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  all  its  parts  taken  together. 

47.  Postulates. 

Let  it  be  granted  — 

1.  That  a  straight  line  can  be  drawn  from  any  one  point  to  any 

other  point. 

2.  That  a  straight  line  can  be  produced  to  any  distance,  or  can 

be  terminated  at  any  point. 

3.  That  the  circumference  of  a  circle  can  be  described  about  any 

centre,  at  any  distance  from  that  centre. 


14 


GEOMETRY. BOOK   I. 


48.  Symbols  and  Abbreviations. 


.'.  therefore. 

=  is  (or  are)  equal  to. 

Z  angle. 

A  angles. 

A  triangle. 

A  triangles. 

II    parallel. 
O  parallelogram 
HJ  parallelograms. 
_L  perpendicular. 
Jl  perpendiculars, 
rt.  Z  right  angle, 
rt.  A  right  angles. 

>  is  (or  are)  greater  than. 
<  is  (or  are)  less  than, 
rt.  A  right  triangle, 
rt.  A  right  triangles. 

O  circle. 

(D  circles. 

+  increased  by. 

—  diminished  by. 

X  multiplied  by. 

-r-  divided  by. 


Post,  postulate. 

Def.  definition. 

Ax.  axiom. 

Hyp.  hypothesis. 

Cor.  corollary. 

Q.  E.  D.  quod  erat  demonstran- 
dum. 

Q.  E.  F.  quod  erat  faciendum. 

Adj.  adjacent. 

Ext.- int.  exterior-interior. 

Alt. -int.  alternate-interior. 

Iden.  identical. 

Cons,  construction. 

Sup.  supplementary. 

Sup.  adj.    supplementary-adja- 
cent. 

Ex.  exercise. 

111.  illustration. 


perpendicular  and  oblique  lines.  15 

On  Perpendicular  and  Oblique  Lines. 

Proposition  I.     Theorem. 

49.    When  one  straight  line  crosses  another  straight  line 
the  vertical  angles  are  equal. 


P 

Let  line  0  P  cross  A  B  at  C. 

We  are  to  prove      Z  OCB  =  Z  A  C  P. 

ZOCA  +  ZOCB  =  2  rt.  A,  §  34 

{fn  ing  siqy.-adj.  A). 

ZOCA  +  ZACP=2vt.  A,  §  34 

(being  sup.  -  adj.  A). 

.'.ZOCA  +  ZOCB  =  ZOCA  +  ZACP.     Ax.  1. 

Take  away  from  each  of  these  equals  the  common  Z  0  C  A. 
Then  ZOCB  =  Z  AC  P. 

In  like  manner  we  may  prove 

Z  ACO  =  Z  PCB. 

Q.  E.  D. 

50.  Corollary.  If  two  straight  lines  cut  one  another,  the 
four  angles  which  they  make  at  the  point  of  intersection  are 
together  equal  to  four  right  angles. 


L6  GEOMETRY. BOOK    I. 


Proposition  II.     Theorem. 

51.  When  the  sum  of  two  adjacent  angles  is  equal  to  two 
right  angles,  their  exterior  sides  form  one  and  the  same 
straight  line. 

0 


--F 


Let  the  adjacent  angles  Z  OCA  +  Z  0  C B  =  2  rt.  A. 
We  are  to  prove  A  C  and  C  B  in  the  same  straight  line. 
Suppose  C  F  to  be  in  the  same  straight  line  with  A  C. 

Then  ZOCA  +  ZOCF=2  it.  A.  §34 

(being  sup. -adj.  A). 

But  ZOCA  +  ZOCB  =  2Tt  A.  Hyp. 

.•.ZOCArt-ZOCF=ZOCA  +  ZOCB.     Ax.  1. 

Take  away  from  each  of  these  equals  the  common  Z  0  C  A. 

Then  Z  OCF=Z  OGB. 

.'.  C  B  and  C F  coincide,  and  cannot  form  two  lines  as  rep- 
resented in  the  figure. 

.'.AC  and  C  B  are  in  the  same  straight  line. 

Q.  E.  D. 


PERPENDICULAR    AND    OBLIQUE    LINES. 


17 


Proposition  III.     Theorem. 
52.  A  perpendicular  measures  the  shortest  distance  from 
a  point  to  a  straight  line. 


Let  A  B  be  the  given  straight  line,  C  the  given  point, 
and  CO  the  perpendicular. 

We  are  to  prove  C 0  <  any  other  line  drawn  from  C  to  A  B, 
as  C  F. 

Produce  CO  to  E,  making  0  E  =  C  0. 

Draw  EF. 
On  A  B  as  an  axis,  fold  over  0  C  F  until  it  comes  into  the 
plane  of  OFF. 

The  line  0  C  will  take  the  direction  of  0  E, 
(since  ZCOF=ZEOFf  each  being  a  rt.  Z. ). 

The  point  C  will  fall  upon  the  point  E, 
(since  0  0—  0  E  by  cons. ). 
.-.line  C  F=  line  F  E}  §  18 

(having  their  extremities  in  tlie  same  points). 

.'.  CF+  FE=2  CF, 
and  CO  +OE=2  CO.  Cons. 

But  CO+OE<CF+FE,  §18 

(a  straight  line  is  tlie  shortest  distance  between  two  points). 
Substitute  2  C  0  for  C  0  +  0  E, 

and         2  CF  for  CF+  F  E ;  then  we  have 
2  C0<2CF. 

.'.  CO  <  CF. 

Q.  E.  D. 


18 


GEOMETRY. BOOK  I. 


Proposition  IV.     Theorem. 

53.  Two  oblique  lines  drawn  from  a  point  in  a  perpen- 
dicular, cutting  off  equal  distances  from  the  foot  of  the  per- 
pendicular, are  equal. 


Let  F  C  be  the  perpendicular,  and  C  A  and  C  0  two 
oblique  lines  cutting  off  equal  distances  from  F. 


We  are  to  prove  C  A  =  C  0. 

Fold  over  C FA,  on  C F  as  an  axis,  until  it  comes  into  the 
plane  of  C  F  0. 

FA  will  take  the  direction  of  FO, 
(since  Z.CFA  =  ZCFO,  each  being  art.  /.). 

Point  A  will  fall  upon  point  0, 
(FA  =  FO,  by  hyp.). 


.-.line  C A  =  line  CO, 
(their  extremities  being  the  same  points). 


§18 


Q.  E.  D. 


PERPENDICULAR    AND    OBLIQUE    LINES.  19 


Proposition  V.     Theorem. 

54.  The  sum  of  two  lines  drawn  from  a  point  to  the  ex- 
tremities  of  a  straight  line  is  greater  than  the  sum  of  two 
other  lines  similarly  drawn,  but  included  by  them. 


A  B 

Let  C  A  and  C  B  be  two  lines  drawn  from  the  point  C 
to  the  extremities  of  the  straight  line  A  B.  Let  0  A 
and  0  B  be  two  lines  similarly  drawn,  but  included 
byCA  andCB. 

We  are  to  prove      CA  +  CB>OA  +  OB. 

Produce  A  0  to  meet  the  line  C  B  at  E. 

Then  AC+  CE>AO+  OB,  §18 

(a  straight  line  is  the  shortest  distance  between  tico  2>oints), 

and  BE+  OE>  BO.  §  18 

Add  these  inequalities,  and  we  have 

CA  +  CE+BE+OE>OA  +  OE+OB. 

Substitute  for  CE  +  BE  its  equal  C B, 

and  take  away  0  E  from  each  side  of  the  inequality. 

We  have  CA  +  CB  >  0  A  +  0  B. 

Q.  E.  D 


20 


GEOMETRY. 


BOOK   I. 


Proposition  VI.     Theorem. 

55.   Of  two  oblique  lines  drawn  from  the  same  point  in  a 
perpendicular,  cutting  off  unequal  distances  from  the  foot  of 
the  perpendicular }  the  more  remote  is  the  greater. 
C 


Let  G F  be  perpendicular  to  A  B,  and  C K  and  C  H  two 
oblique  lines  cutting  off  unequal  distances  from  F. 

We  are  to  prove  C  H  >  C  K. 

Produce  C F  to  E,  making  FE=CF. 
Draw  EK  and  EH. 

CH=HE,?m&CK=  KE,  §53 

(two  oblique  lines  drawn  from  the  same  point  in  a  J_,  cutting  off  equal  dis- 
tances from  the  foot  of  the  _L,  are  equal). 

But  CH+HE>CK+KE,  §54 

(The  sum  of  two  oblique  lines  drawn  from  a  point  to  the  extremities  of  a 
straight  line  is  greater  than  tht  sum  of  two  other  lines  similarly  drawn, 
but  included  by  them); 

.\2  CII>2CK; 

.\CH>CK. 

Q.  E.  D. 

56.  Corollary.  Only  two  equal  straight  lines  can  be  drawn 
from  a  point  to  a  straight  line ;  and  of  two  unequal  lines,  the 
greater  cuts  off  the  greater  distance  from  the  foot  of  the  perpen- 
dicular. 


PERPENDICULAR   AND    OBLIQUE    LINES. 


21 


Proposition  VII.     Theorem. 


57.  Two  equal  oblique  lines,  drawn  from  the  same  point 
in  a  perpendicular,  cut  off  equal  distances  from  the  foot  of 
the  perpendicular. 


.  C 


Let  C  F  be  the  perpendicular,  and  C  E  and  C  K  be  two 
equal  oblique  lines  drawn  from  the  point  C. 


We  are  to  prove 


FE=FK.< 


Fold  over  C  FA  on  C  F  as  an  axis,  until  it  comes  into  the 
plane  of  CFB. 

The  line  FE  will  take  the  direction  FK, 
(Z  CFE  =  ZCFK,  each  being  a  rt.  Z  ). 

Then  the  point  E  must  fall  upon  the  point  K ; 

otherwise  one  of  these  oblique  lines  must  be  more  remote  from 
the  _L, 

and  .'.  greater  than  the  other;    which  is  contrary  to  the 
hypothesis.  §  55 


.'.FE  =  FK. 


Q.  E.  D. 


GEOMETRY.  —  BOOK    I. 


Proposition  VIII.     Theorem. 

58.  If  at  the  middle  point  of  a  straight  line  a  perpen- 
dicular be  erected, 

I.  Any  point  in  the  perpendicular  is  at  equal  distances 
from  the  extremities  of  the  straight  line. 

II.  Any  point  without  the  perpendicular  is  at  unequal 
distances  from  the  extremities  of  the  straight  line. 


Let  P R  be  a  perpendicular  erected  at  the  middle  01 
the  straight  line  A  B,  0  any  point  in  PR,  and  C  any 
point  without  P R. 

I.  Draw  OA  and  OB. 
We  are  to  prove  0  A  =  0  B. 
Since                           P  A  =  P  B, 

OA  =  OB,  §  53 

(two  oblique  lines  drawn  from  tlie  same  jmint  in  a  ±,  cutting  off  equal  dis- 
tances  from  tJw,  foot  of  the  ±,  are  equal). 

II.  Draw  CA  and  C  B. 

We  are  to  prove  C  A  and  C  B  unequal. 

One  of  these  lines,  as  CA,  will  intersect  the  _L. 
From  D,  the  point  of  intersection,  draw  D  B. 


PERPENDICULAR   AND    OBLIQUE    LINES.  23 

DB  =  DA,  §53 

(five  oblique  lines  drawn  from  the  same  point  in  a  ±,  cutting  off  equal  dis- 
tances from  the  foot  of  the  ±,  are  equal). 

CB<  CD  +  I)B,  §  18 

(a  straigld  line  is  Uie  shortest  distance  between  two  points). 

Substitute  for  D  B  its  equal  D  A,  then 

CB<  CD  +  DA. 

But  CD  +  DA  =  CA,  Ax.  9. 

,'.CB<  CA. 

Q.  E.  D. 

59.  The  Locus  of  a  point  is  a  line,  straight  or  curved,  con- 
taining all  the  points  which  possess  a  common  property. 

Thus,  the  perpendicular  erected  at  the  middle  of  a  straight 
line  is  the  locus  of  all  points  equally  distant  from  the  extremi- 
ties of  that  straight  line. 

60.  Scholium.  Since  two  points  determine  the  position  of 
a  straight  line,  two  points  equally  distant  from  the  extremities 
of  a  straight  line  determine  the  perpendicular  at  the  middle 
point  of  that  line. 


Ex.  1.    If  an  angle  be  a  right  angle,  what  is  its  complement? 

2.  If  an  angle  be  a  right  angle,  what  is  its  supplement  1 

3.  If  an  angle  be  #  of  a  right  angle,  what  is  its  complement  1 

4.  If  an  angle  be  £  of  a  right  angle,  what  is  its  supplement  1 

5.  Show  that  the  bisectors  of  two  vertical  angles  form  one 
and  the  same  straight  line. 

6.  Show  that  the  two  straight  lines  which  bisect  the  two 
pairs  of  vertical  angles  are  perpendicular  to  each  other. 


24 


GEOMETRY.  —  BOOK   I. 


Proposition  IX.     Theorem. 

61.  At  a  point  in  a  straight  line  only  one  perpendicular 
to  that  line  can  be  drawn ;  and  from  a  point  without  a 
straight  line  only  one  perpendicular  to  that  line  can  be  drawn. 


AE 


A  F 


B 

Fig.  1. 


EB 

Fig.  2. 


Let  B  A  {fig.  1)  be  perpendicular  to  C  D  at  the  point  B. 

We  are  to  prove  B  A  the  only  p>erpendicular  to  G  D  at  the 
point  B. 

If  it  be  possible,  let  B  E  be  another  line  _L  to  G  D  at  B. 

Then  Z  EBD  is  a  rt.  Z.  §26 

But  Z  ABD  is  art.  Z.  §  26 

.'.Z  EBD  =  Z  ABD.  Ax.  1. 

That  is,  a  part  is  equal  to  the  whole ;  which  is  impossible. 

In  like  manner  it  may  be  shown  that  no  other  line  but  B  A 
is  _L  to  GD  at  B. 

Let  AB  {fig.  2)  be  perpendicular  to  G D  from  the  point  A. 
We  are  to  prove  A  B  the  only  _L  to  G  D  from  the  point  A. 

If  it  be  possible,  let  A  E  be  another  line  drawn  from  A  1_ 
to  GD. 

Conceive  Z  A  E  B  to  be  moved  to  the  right  until  the  ver- 
tex E  falls  on  B,  the  side  E  B  continuing  in  the  line  G  D. 

Then  the  line  E  A  will  take  the  position  B  F. 

Now  if  A  E  be  J_  to  C  D,  B  F  is  JL  to  C  D,  and  there  will 
be  two  J»  to  C  D  at  the  point  B ;  which  is  impossible. 

In  like  manner,  it  may  be  shown  that  no  other  line  but 
A  B  is  _L  to  GD  from  A.  q.  e.  d. 

62.  Corollary.  Two  lines  in  the  same  plane  perpendicular 
to  the  same  straight  line  have  the  same  direction  ;  otherwise 
they  would  meet  (§  22),  and  we  should  have  two  perpendicular 
lines  drawn  from  their  point  of  meeting  to  the  same  line ;  which 
is  impossible. 


PARALLEL  LINES.  25 


On  Parallel  Lines. 

63.  Parallel  Lines  are  straight  lines  which  lie  in  the  same 
plane  and  have  the  same  direction,  or  opposite  directions. 

Parallel  lines  lie  in  the  same  direction,  when  they  are  on 
the  same  side  of  the  straight  line  joining  their  origins. 

Parallel  lines  lie  in  opposite  directions,  when  they  are  on 
opposite  sides  of  the  straight  line  joining  their  origins. 

64.  Two  parallel  lines  cannot  meet.  §  21 
65.'    Two  lines  in  the  same  plane  perpendicular  to  a  given 

line  have  the  same  direction  (§  62),  and  are  therefore  parallel. 

66.  Through  a  given  point  only  one  line  can  be  drawn  par- 
allel to  a  given  line.  §  18 


If  a  straight  line  EF  cut  two  other  straight  lines  A  B 
and  C  D,  it  makes  with  those  lines  eight  angles,  to  which  par- 
ticular names  are  given. 

The  angles  1,  4,  6,  7  are  called  Interior  angles. 

The  angles  2,  3,  5,  8  are  called  Exterior  angles. 

The  pairs  of  angles  1  and  7,  4  and  6  are  called  Alternate- 
interior  angles. 

The  pairs  of  angles  2  and  8,  3  and  5  are  called  Alternate- 
exterior  angles. 

The  pairs  of  angles  1  and  5,  2  and  6,  4  and  8,  3  and  7  are 
called  Exterior-iyiterior  angles. 


GEOMETRY. BOOK   I. 


Proposition  X.     Theorem. 

67.    If  a  straight  line  be  perpendicular  to  one  of  two 
parallel  lines }  it  is  perpendicular  to  the  other. 


M> 

E- 


~~-X 


Let  A  B  and  E F  be  two  parallel  lines,  and  let  H K  be 
p erp  en  die  ular  to  A  B. 

We  are  to  prove      H  K  _L  to  E  F. 

Through  C  draw  UN  JL  to  UK. 

Then  '  MN  is  II  to  A  B.  §  65 

{Two  lines  in  the  same  plane  JL  to  a  given  line  arc  parallel). 

But  EFi&WtoAB,  By  p. 

.'.  E  F  coincides  with  M N.  §  G6 

{Through  the  same  point  only  one  line  can  be  drawn  \\  to  a  given  line). 

.'.  E  F  is  ±  to  HK, 
that  is  UK  is -L  to  EF. 

Q.  E.  D. 


PARALLEL   LINES.  27 


Proposition  XI.     Theorem. 

68.   If  two  parallel  straight  lines  he  cut  by  a  third 
straight  line  the  alternate-interior  angles  are  equal. 

A  B F 


c         b H 

Let  E  F  and  Gil  be  two  parallel  straight  lines  cut  by 
the  line  BO. 

We  are  to  prove  Z  B  =  /.0. 

Through  0,  the  middle  point  of  B C,  draw  A  D ±.to  G H. 

Then  A  D  is  likewise  _L  to  E  F,  §  67 

(a  straight  line  ±  to  one  of  two  lis  is  ±  to  the  oilier), 

that  is,  C  D  and  B  A  are  both  _L  to  A  D. 
Apply  figure  C  0  D  to  figure  B  0  A  so  that  0  D  shall  fall 
on  OA. 

Then  0  C  will  fall  on  OB, 

(since  Z  CO  D  =  /.BOA,  being  vertical  A)  ; 

and  point  C  will  fall  upon  B, 

(siiice  0  C  —  0  B  by  construction). 

Then        _L  CD  will  coincide  with  ±  B A,  §  61 

(/rem  a  point  without  a  straig/U  line  only  one  ±  to  tliat  line  can  be  drawn). 

.'.  Z.  0  G D  coincides  with  Z  0  B  A,  and  is  equal  to  it. 

Q.  E.  D. 

Scholium.  By  the  converse  of  a  proposition  is  meant  a 
proposition  which  has  the  hypothesis  of  the  first  as  conclusion 
and  the  conclusion  of  the  first  as  hypothesis.  The  converse  of 
a  truth  is  not  necessarily  true.  Tims,  parallel  lines  never  meet ; 
its  converse,  lines  which  never  meet  are  parallel,  is  not  true  unless 
the  lines  lie  in  the  same  plane. 


Note.  —  The  converse  of  many  propositions  will  be  omitted, 
but  their  statement  and  demonstration  should  be  required  as  an 
important  exercise  for  the  student. 


28  GEOMETRY.  —  BOOK  I. 


Proposition  XII.     Theorem. 

69.  Conversely  :  When  two  straight  lines  are  cut  by  a 
third  straight  line,  if  the  alternate-interior  angles  be  equal, 
the  two  straight  lines  are  parallel. 


Let  E  F  cut  the  straight  lines  A  B  and  C  D  in  the  points 
H  and  K,  and  let  the  Z  A  HK  =  Z  HKD. 

We  are  to  prove      A  B  II  to  C  D. 
Through  the  point  H  draw  M N  II  to  CD ; 

then  Z  MHK  =  Z  H K D,  §  68 

(being  alt.  -int.  A  ). 

But  Z  A  HK  =  Z  HKD,  Hyp. 

.'.Z  MHK=  Z  AHK.  Ax.  1. 

.'.the  lines  M  N  and  A  B  coincide. 
But  J/iVis  II  to  CD;  Cons. 

.'.  AB,  which  coincides  with  M N,  is  II  to  CD. 

Q.  E.  o. 


PARALLEL   LINES.  29 


Proposition  XIIT.     Theorem. 

70.   If 'two parallel  lines  be  cut  by  a  third  straight  line, 
the  exterior-interior  angles  are  equal. 

E 


Let  AB  and  C  D   be  two  parallel  lines   cut   by  the 
straight  line  E  F,  in  the  points  II  and  K. 

We  are  to  prove      Z  EHB  =  Z  H K D. 

ZEHB  =  /.AIIK,  §49 

{being  vertical  A). 

But  ZAIIK  =  ZIIKD,  §  08 

(being  alt. -int.  A). 

.-.Z  EIIB  =  Z  HKD.  Ax.  1 

In  like  manner  we  may  prove 

ZEIIA  =  ZHKC. 

Q.  E.  D. 


71.  Corollary.    The  alternate-exterior  angles,  EHB  and 
C  K  F,  and  also  A  II E  and  D  K  F}  are  equal. 


30  GEOMETRY.  —  BOOK   I. 


Proposition  XIV.     Theorem. 

72.  Conversely  :  When  two  straight  lines  are  cut  by  a 
third  straight  line,  if  the  exterior-interior  angles  be  equal, 
these  two  straight  lines  are  parallel. 


Let  EF  cut   the  straight  lines  AB  and  CD  in   the 
points  II  and  K,  and  let  the  Z  EHB  =  Z  HKD. 

We  are  to  prove      A  B  II  to  C  D  . 
Through  the  point  H  draw  the  straight  line  M  N  II  to  CD. 

Then  ZEHN=ZHKD,  §70 

(being  ext.  -int.  A  ). 

But  Z  EHB  =  Z  HKD.  Hyp. 

.-.  Z  EHB  =  Z  EHN.  Ax.  1. 

.*.  the  lines  M N  and  A  B  coincide. 
But  MNh  II  to  CD,  Cons. 

,\  AB,  which  coincides  with  M N,  is  II  to  CD. 

Q.  E.  D. 


PARALLEL    LINES.  31 


Proposition  XV.     Theorem. 

73.  If  two  parallel  lines  be  cut  by  a  third  straight  line, 
the  sum  of  the  two  interior  angles  on  the  same  side  of  the 
secant  line  is  equal  to  two  right  angles. 

E 


Let  AB  and  C  D  be    two  parallel  lines   cut   by   the 
straight  line  EF  in  the  points  II  and  K. 

We  are  to  prove      A  B II K  +  Z  II K  D  =  two  rt.  A. 

Z  EIIB  +  Z  BIIK=2  rt.  A,  §  34 

(being  sup.  -adj.  A  ). 

But  Z  EIIB  =  A  II KB,  §70 

(being  exl.-int.  A  ). 

Substitute  Z  HKD  for  Z  EIIB  in  the  first  equality; 
then  Z  BHK  +  Z  HKD  =  2  rt.  A. 

Q.  E.  d. 


32  GEOMETRY.  —  BOOK   I. 

Proposition  XVI.     Theorem. 

74.  Conversely  :  When  two  straight  lines  are  cut  hy  a 
third  straight  line,  if  the  two  interior  angles  on  the  same  side 
of  the  secant  line  he  together  equal  to  two  right  angles,  then 
the  two  straight  lines  are  parallel. 

E 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the 
points  H  and  K,  and  let  the  Z.  B  II K  +  A  H  K D 
equal  two  right  angles. 

We  are  to  prove      A  B  II  to  CD. 

Through  the  point  IT  draw  MN  II  to  CD. 

Then         Z.  NHK  +  Z  HKD  =  2  rt.  A,  §  73 

(being  two  interior  A  on  the  same  side  of  the  secant  line). 

But  ZBHK+ZHKD  =  2  rt.  A.  Hyp. 

.'./.NHK+AHKD  =  ABHK+/.HKD.  Ax.  1. 
Take  away  from  each  of  these  equals  the  common  /.HKD, 
then  ANHK=ABHK. 

.'.  the  lines  A  B  and  M N  coincide. 
But  MN\%  II  to  CD;  Cons. 

.',  AB.  which  coincides  with  M iV,  is  II  to  CD. 

Q.  E    D. 


PARALLEL    LINES. 


83 


Proposition  XVII.     Theorem. 

75.  Two  straight  lines  which  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 

H 

i 


A' 
Let  AB  and  CD  be  parallel  to  E  F. 
We  are  to  prove      A  B  II  to  C D. 

Draw  H K±  to  EF. 

Since  C  D  and  EF  are  II,  HK  is  J_  to  C  D,  §  67 

(if  a  straigld  line  be  _L  to  one  of  tiro  lis,  it  is  _L  to  the  other  also). 

Since  A  B  and  EF  are  II,  UK  is  also  _L  to  A  B,       §  67 

.\ZHOB  =  Z  II PD, 

(each  being  art.  /.). 

.'.  ABisW  to  CI),  §  72 

(ivhen  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  ext.  -int.  A 
be  equal,  the  two  Hues  are  II  ). 

Q.  E.  D. 


34 


GEOMETRY. 


BOOK    I. 


Proposition  XVIII.     Theorem. 

76.   Two  parallel  lines  are  everywhere  equally  distant 
from  each  other. 

E  M  H 


I) 


F  P  K 

Let  A  B  and  CD  be  two  parallel  lines,  and  from  any 
two  points  in  A  B,  as  E  and  II,  let  EF  and  II K 
be  drawn  perpendicular  to  A  B. 
We  are  to  prove      E F =  UK 

Now  EF  and  UK  are  J_  to  C  D,  §  67 

(a  line  ±  to  one  of  two  lis  is  ±  to  the  other  also). 

Let  M  be  the  middle  point  of  E  H. 
*     Draw  MP  ±  to  A  B. 
On  MP  as  an  axis,  fold  over  the  portion  of  the  figure  on 
the  right  of  MP  until  it  comes  into  the  plane  of  the  figure  on 
the  left. 

MB  will  fall  on  MA, 
(for  ZPMH=APME,  each  being  art.  Z )  ; 

the  point  H  will  fall  on  E, 
{for  M H=  ME,  by  hyp.)  ; 

HK  will  fall  on  EF, 

(for  ZMHK=  ZMEF,  each  being  art.  Z )  ; 

and  the  point  K  will  fall  on  E  F,  or  E  F  produced. 

Also,  PD  will  fall  on  P  C, 

(Z  MPK=  Z  MPF',  each  being  a  rt.  Z)  ; 

and  the  point  K  will  fall  on  P  C. 

Since  the  point  K  falls  in  both  the  lines  EF  and  P  C, 

it  must  fall  at  their  point  of  intersection  F. 

.\HK=  EF,  §  18 

(their  extremities  being  the  same  points). 

Q.  E.  D. 


PARALLEL    LINES. 


35 


Proposition  XIX.     Theorem. 

77.   Two  angles  whose  sides  are  parallel,  two  and  two, 
and  lie  in  the  same  direction,  or  opposite  directions,  from  their 
vertices,  are  equal. 
A        D 


D> 

Fig.  1.  Fig.  2. 

Let  A  B  and  E  (Fig.  1)  have  their  sides  B  A  and  E  D, 
and  BC  and  EF  respectively,  parallel  and  lying 
in  the  same  direction  from  their  vertices. 

We  are  to  prove  the      Z  B  =  Z  E. 

Produce  (if  necessary)  two  sides  which  are  not  II  until  they 

intersect,  as  at  H ; 

then  Z  B  =  Z  DHC,  '  §  70 

(being  i.r/.-inf.  A ), 

and  ZE  =  ZDHC,  §70 

.'.ZB  =  ZE.  Ax.  1 

Let  A  B'  and  E>  (Fig.  2)  have  B'  A1  and  W  D',  and  B'  C 
and  E'  F'  respectively,  parallel  and  lying  in  oppo- 
site directions  from  their  vertices. 

We  are  to  prove  the       Z  B'  =  Z  E1. 
Produce  (if  necessary)  two  sides  which  are  not 
intersect,  as  at  H1. 

Then  Z  B'  =  Z  E  IT  C, 

[I"  in'/  i  xt. -inf.  A), 


and 


Z  E'  =  Z  E>  H'  C 

(beiiuj  alt. -int.  A  )  ; 

.'.  Z  B'  =  Z  E> 


until  they 
§70 
§68 


Ax.  1. 

Q.  E.  D. 


36  GEOMETRY. BOOK    I. 


Proposition  XX.     Theorem. 


78.  If  two  angles  have  two  sides  parallel  and  lying  in 
the  same  direction  from  their  vertices,  while  the  other  two 
sides  are  parallel  and  lie  in  opposite  directions,  then  the  two 
angles  are  supplements  of  each  other. 


Let  A  BC  and  D  E  F  be  two  angles  having  B C  and  ED 
parallel  and  lying  in  the  same  direction  from  their 
vertices,  while  E F  and  B  A  are  parallel  and  lie  in 
opposite  directions. 

We  are  to  prove  /.ABC  and  Z  D  E  F  supplements  of  each 
other. 

Produce  (if  necessary)  two  sides  which  are  not  II  until  they 
intersect  as  at  H. 

ZABC  =  ZBHD,  §70 

(being  ext.-int.  A  ). 

ZDEF==ZBHE,  §68 

(being  alt.  -int.  A  ). 

But  Z  B II D  and  Z  B  HE  are  supplements  of  each  other,     §  34 

sup. -adj.  A ). 


.'.  Z  ABC  and  Z  D  E  F,   the  equals  of  Z  BED  and 
Z  B  H  E,  are  supplements  of  each  other. 


Q.  E.  D. 


TRIANGLES.  37 


On  Triangles. 

79.  Def.  A  Triangle  is  a  plane  figure  bounded  by  three 
straight  lines. 

A  triangle  has  six  parts,  three  sides  and  three  angles. 

80.  When  the  six  parts  of  one  triangle  are  equal  to  the  six 
parts  of  another  triangle,  each  to  each,  the  triangles  are  said  to 
be  equal  in  all  respects. 

81.  Def.  In  two  equal  triangles,  the  equal  angles  are  called 
Homologous  angles,  and  the  equal  sides  are  called  Homologous 
sides. 

82.  In  equal  triangles  the  equal  sides  are  opposite  the 
equal  angles. 


ISOSCELES.  EQUILATERAL. 


83.  Def.    A  Sealene  triangle  is  one  of  which  no  two  sides 
are  equal. 

84.  Def.    An  Isosceles  triangle  is  one  of  which  two  sides 
<ire  equal. 

85.  Def.    An  Equilateral  triangle  is  one  of  which  the  three 
sides  are  equal. 

86.  Def.    The  Base  of  a  triangle  is  the  side  on  which  the 
triangle  is  supposed  to  stand. 

In  an  isosceles  triangle,  the  side  which  is  not  one  of  the 
equal  sides  is  considered  the  base. 


38 


GEOMETRY. 


BOOK    I. 


87.  Def.    A  Right  triangle  is  one  which   has  one  of  the 
angles  a  right  angle. 

88.  Def.    The  side  opposite  the  right  angle  is  called  the 
Hypotenuse. 

89.  Def.    An  Obtuse  triangle  is  one  which  has  one  of  the 
angles  an  obtuse  angle. 

90.  Def.    An  Acute  triangle  is  one  which  has  all  the  angles 
acute. 


EQUIANGULAR. 


91.  Def.  An  Equiangular  triangle  is  one  which  has  all 
the  angles  equal. 

92.  Def.  In  any  triangle,  the  angle  opposite  the  base  is 
called  the  Vertical  angle,  and  its  vertex  is  called  the  Vertex  of 
the  triangle. 

93.  Def.  The  Altitude  of  a  triangle  is  the  perpendicular 
distance  from  the  vertex  to  the  base,  or  the  base  produced. 

94.  Def.  The  Exterior  angle  of  a  triangle  is  the  angle  in- 
cluded between  a  side  and  an  adjacent  side  produced,  as  /.  CBD. 

95.  Def.  The  two  angles  of  a  triangle  which  are  opposite 
the  exterior  angle,  are  called  the  two  opposite  interior  angles,  as 
A  A  and  G. 


-in  ANGLES.  39 


96.  Any  side  of  a  triangle  is  less  than  the  sum  of  the 
other  two  sides. 

Since  a  straight  line  is  the  shortest  distance  between  two 

points, 

AC<AB+BC 

97.  Am/  tide  of  a  triangle  is  greater  than  the  difference 
of  the  other  two  sides. 

In  the  inequality  A  C  <  A  B  +  B  C, 

take  away  A  B  from  eacli  side  of  the  inequality. 

Then  AC~AB<BC;  or 

BO  AG-  AB. 


Ex.  1.  Show  that  the  sum  of  the  distances  of  any  point  in  a 
triangle  from  the  vertices  of  three  angles  of  the  triangle  is  greater 
than  half  the  sum  of  the  sides  of  the  triangle. 

2.  Show  that  the  locus  of  all  the  points  at  a  given  distance 
from  a  given  straight  line  A  B  consists  of  two  parallel  lines, 
drawn  on  opposite  sides  of  A  B,  and  at  the  given  distance 
from  it. 

3.  Show  that  the  two  equal  straight  lines  drawn  from  a  point 
to  a  straight  line  make  equal  acute  angles  with  that  line. 

4.  Show  that,  if  two  angles  have  their  sides  perpendicular, 
each  to  each,  they  are  either  equal  or  supplementary. 


40  GEOMETRY. BOOK    I. 


Proposition  XXI.     Theorem. 

98.   The  sum  of  the  three  angles  of  a  triangle  is  equal 
to  two  right  angles. 


C ' 

Let  A  B  G  be  a  triangle. 

We  are  to  prove     AB-\rABCA  +  AA  =  two  rt.  A. 

Draw  C  E  II  to  A  B,  and  prolong  A  C. 

ThenZ  ECF+ZECB+ZBCA  =  2  rt.  A,  §  34 
(the  sum  of  all  the  A  about  a  point  on  tlu  same  side  of  a  straight  line 

=  2rt.  A ). 

But  ZA=ZECF,  §70 

(being  ext.  -int.  A ), 

and  Z  B  =  Z.  BCE,  §68 

( being  alt.  -int.  A  ) . 

Substitute  for  Z  E  CF&nd  Z  B  C  E  their  equal  A,  A  and  B. 
Then        Z  A  +  Z  B  +  Z  B  OA  =  2  rt.  A. 

Q.  E.  D. 

99.  Corollary  1.  If  the  sum  of  two  angles  of  a  triangle  be 
known,  the  third  angle  can  be  found  by  taking  this  sum  from 
two  right  angles. 

100.  Cor.  2.  If  two  triangles  have  two  angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angles  will  be  equal. 


TRIANGLES.  41 


101.  Cor.  3.  If  two  right  triangles  have  an  acute  angle 
of  the  one  equal  to  an  acute  angle  of  the  other,  the  other  acute 
angles  will  be  equal. 

102.  Cor.  4.  In  a  triangle  there  can  be  but  one  right  angle, 
or  one  obtuse  angle. 

103.  Cor.  5.  In  a  right  triangle  the  two  acute  angles  are 
complements  of  each  other. 

104.  Cor.  6.  In  an  equiangular  triangle,  each  angle  is  one 
third  of  two  right  angles,  or  two  thirds  of  one  right  angle. 


Proposition  XXII.     Theorem. 

105.   The  exterior  angle  of  a  triangle  is  equal  to  the  sum 
of  the  two  opposite  interior  angles. 


Let  BC II  be  an  exterior  angle  of  the  triangle  ABC. 
We  are  to  prove      Z  B  Gil  =  Z  A  +  Z  B. 

Z  BCII  +  Z  ACB=2vt  A,  §  34 

(beivcf  sup.-mlj.  A). 

Z  A  +  Z  B+  Z  ACB  =  2  rt.  A,  §  98 

{three  AofaA  =  two  rt.  A  ). 

.'.Z  BCII+  Z  ACB  =  Z  A  +  Z  B  +  Z  A  C B.      Ax.  1. 

Take  away  from  each  of  these  equals  the  common  Z  A  C  B ; 

then  ZBCII  =  ZA  +  ZB. 

Q.  E-  D. 


42 


GEOMETRY. BOOK    I. 


Proposition  XXIII.     Theorem. 

106.  Two  triangles  are  equal  in  all  respects  when  two 
sides  and  the  included  angle  of  the  one  are  equal  respectively 
to  two  sides  and  the  included  angle  of  the  other. 


B   A' 


In    the    triangles   ABO   and  A' B' C,   let  AB  =  A'h', 

A  $=A'  C',Z  A=Z  A'. 

We  are  to  prove      A  A  B  C  =  A  A1  B'  C. 

Take  up  the  A  A  B  C  and  place  it  upon  the  A  A'  B'  C  so 
that  A  B  shall  coincide  with  A'  B'. 


Then        A  G  will  take  the  direction  of  A'  C, 
(for  Z  A  =  Z  A',  by  hyjy.), 

the  point  C  will  fall  upon  the  point  C, 
(forAC=A'Ci,byhyp.); 

.\CB  =  C  B', 

(their  extremities  being  the  same  points). 

,'.  the  two  A  coincide,  and  are  equal  in  all  respects. 


IS 


Q.  E.  D, 


TRIANGLES.  43 


Proposition  XXIV.     Theorem. 

107.  Two  triangles  are  equal  in  all  respects  when  a  side 
and  two  adjacent  angles  of  the  one  are  equal  respectively  to  a 
side  and  two  adjacent  angles  of  the  other. 

C  C 


B  A' 


In   the   triangles   A  BC   and  A'  B'  C,   let  A  B  =  A'  B', 
Z  A  =  Z  A',  Z  B  =  Z  B'. 

We  are  to  prove      A  A  B  C  =  A  A'  B'  C. 

Take  up  A  A  BC  and  place  it  upon  A  A'  B' C,   so  that 
A  B  shall  coincide  with  A'  B'. 

A  C  will  take  the  direction  of  A'  C, 
(for  Z  A  =  ZA',  by.  hyp.) ; 

the  point  C,  the  extremity  of  A  C,  will  fall  upon  A'  C  or 
A'  C  produced. 

B  0  will  take  the  direction  of  B1  C, 
(for  ZB  =  ZB',  by  hyp.); 

the  point  C,  the  extremity  of  B  C,  will  fall  upon  B'  C  or 
B1  C  produced. 

.*.  the  point  C,  falling  upon  both  the  lines  A'  C  and  B'  C, 
must  fall  upon  a  point  common  to  the  two  lines,  namely,  C. 

.*.  the  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


44 


GEOMETRY. 


BOOK    I. 


Proposition   XXV.      Theorem. 

108.   Two  triangles  are  equal  when  the  three  sides  of  the 
one  are  equal  respectively  to  the  three  sides  of  the  other. 
B  B> 


V 

B> 
In   the   triangles  ABC   and  A'  B'  C,  let  A  B  =  A'  B'} 
A  C  =  A'  C,  BG  =  B'C. 
We  are  to  prove      A  A  B  G  =  A  A'  B'  C. 
Place  A  A'  B'  C  in  the  position  A  B'  C,  having  its  greatest 
side  A'  C  in  coincidence  with  its  equal  A  C,  and  its  vertex  at 
B',  opposite  B. 

Draw  B  B'  intersecting  A  C  at  H. 

Since  AB  =  AB',  Hyp. 

point  A  is  at  equal  distances  from  B  and  B'. 

Since  B  C  =  B>  C,  Hyp. 

point  C  is  at  equal  distances  from  B  and  B'. 

.*.  A  C  is  JL  to  BB'  at  its  middle  point,  §  60 

{two  points  at  equal  distances  from  the  extremities  of  a  straight  line  deter- 
mine the  _l_  at  the  middle  of  that  line). 

Now  if  A  A  B'  C  be  folded  over  on  A  G  as  an  axis  until  it 
comes  into  the  plane  of  A  ABC, 

II B'  will  fall  on  H  B, 
(for  /.AHB  =  ZAHB',  each  being  a  rt.  Z), 

and  point  B'  will  fall  on  B, 
(for  HW  =  HB). 

.'.  the  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


TRIANGLES. 


45 


Proposition   XXVI.      Theorem. 

109.  Two  right  triangles  are  equal  when  a  side  and  the 
hypotenuse  of  the  one  are  equal  respectively  to  a  side  and  the 
hypotenuse  of  the  other. 


C     B* 


In  the  right  triangles  ABC  and  A'  B'  C",  let  AB  =  A'  B', 
and  AC  =  A'C. 

We  are  to  prove      A  A  B  C  =  A  A'  B'  C. 

Take  up  the  A  A  B  C  and  place  it  upon  A  A'  B'  C",  so  that 
A  B  will  coincide  with  A'  B'. 

Then  B  C  will  fall  upon  B'  C, 

{for  ZABC=ZA'B'C,  each  being  a  rt.  Z ), 

and  point  C  will  fall  upon  C ; 

otherwise  the  equal  oblique  lines  A  C  and  A'  C  would  cut 
off  unequal   distances    from    the  foot  of  the  _L,   which  is  im- 
possible, §  57 
(two  equal  oblique  lines  from  a  point  in  a  JL  cut  off  equal  distances  from  the 
foot  of  the.  _L). 

,\  the  two  A  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 


46 


GEOMETRY. 


BOOK    I. 


Proposition   XXVII.      Theorem. 


110.  Two  right  triangles  are  equal  token  the  hypotenuse 
and  an  acute  angle  of  the  one  are  equal  respectively  to  the 
hypotenuse  and  an  acute  angle  of  the  other. 


In  the  right  triangles  ABC  and  A'  B'  C,  let  AG  =  A'  C, 
and  Z  A=  Z  A'. 


We  are  to  prove 

AABC  =  AA'B'C. 

AC  =  A'  C, 

Hyp. 

z  a=z  a; 

Hyp. 

then  Z  C  =  Z  C,  \  §  101 

(if  two  rt.  A  have  an  acute  Z  of  the  one  equal  to  an  acute  A  of  the  other, 
then  the  other  acute  A  are  equal). 

.'.AABC  =  AA'B'C,  §  107 

(two   A  are  equal  when  a  side   and  two  adj.   A  of  the  one  are  equal 
respectively  to  a  side  and  two  adj.  A  of  the  other). 

Q.  E.  D. 


111.  Corollary.  Two  right  triangles  are  equal  when  a 
side  and  an  acute  angle  of  the  one  are  equal  respectively  to  an 
homologous  side  and  acute  angle  of  the  other. 


TRIANGLES.  47 


Proposition  XXYIII.   Theorem. 

112.   In  an   isosceles  triangle  the  angles  opposite  the 
equal  sides  are  equal. 

C 


Let  ABC  be  an  isosceles  triangle,  having  the  sides 
AC  and  CB  equal. 

We  are  to  prove      Z  A  =  Z  B. 

From  C  draw  the  straight  line  CE  so  as  to  bisect  the 
Z  A  CB. 

Id  the  A  ACE  and  BCE, 

AC=BCf  Hyp. 

CE=C  E,  Iden. 

ZACE=ZBCE;  Cons. 

.'.AACE  =  ABCE,  §106 

(two  &  are  equal  wlicn  tin,  sides  and  the  included  Z.  of  the  one  are  equal 
respectively  to  two  sides  and  the  included  Z.  of  the  other). 


.'.ZA=ZB, 

(being  liomologous  A  of  equal  A  ). 


Q.  E.  D. 


Ex.  If  the  equal  sides  of  an  isosceles  triangle  be  produced, 
show  that  the  angles  formed  with  the  base  by  the  sides  produced 
are  equal. 


48 


GEOMETRY. BOOK  I. 


Proposition  XXIX.     Theorem. 

113.  A  straight  line  which  bisects  the  angle  at  the  vertex 
af  an  isosceles  triangle  divides  the  triangle  into  two  equal 
triangles,  is  perpendicular  to  the  base,  and  bisects  the  base. 

C 


Let  the  line  C  E  bisect  the  A  AC B  of  the  isosceles 
AACB. 

We  are  to  prove      I.  AACE  =  ABCE; 
II.  UneCE ±to  AB; 
III.  AE  =  BE. 

I.   In  the  A  ACE  and  B  C  E, 

AC=BC,  Hyp. 

CE=CE,  Iden. 

AACE=  ZBCE.  Cons. 

.'.A  ACE  =  A  BCE,  §106 

(having  two  sides  and  the  included  A  of  the  one  equal  respectively  to  two  sides 
and  the  included  A  of  the  other). 

Also,  II.  A  CEA  =  Z  CEB, 

(being  homologous  A  of  equal  A ). 

.*.  CEis±to  AB, 
(a  straight  line  meeting  another,   making  the  adjacent  A  equal,  is  JL  to 

that  line). 

Also,  III.  AE=EB, 

(being  homologous  sides  of  equal  & ). 

Q.  E.  D. 


TRIANGLES. 


49 


Proposition  XXX.     Theorem. 

114.  If  two  angles  of  a  triangle  be  equal,  the  sides  op- 
posite the  equal  angles  are  equal,  and  the  triangle  is  isosceles. 


In  the  triangle  ABC,  let  the  Z  B  =  Z  C. 
We  are  to  prove      AB  =  AC. 

Draw4Z)J_to  BC. 

In  the  rt.  A  A  DB  and  A  D  C, 

AD  =  AD, 

ZB  =  ZC, 

.'.  rt.  A  A  D  B  =  rt.  A  A  D  C, 


Iden. 


§  HI 

(having  a  side  and  an  acute  Z  of  the  one  equal  respectively  to  a  side  and  an 
acute  Z  of  the  other). 

.\AB  =  AC, 
(being  homologous  sides  of  equal  &). 

Q.  E.  D. 


Ex.    Show  that  an  equiangular  triangle  is  also  equilateral. 


50  GEOMETRY.  —  BOOK   I. 

Proposition  XXXI.     Theorem. 

115.  If  two  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  but  the  included  a?igle 
of  the  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  will  be  greater  than  the  third  side 
of  the  second. 

B  B 


E 

In  the  A  ABC  and  ABE,  let  A  B  =  A  B,   BG^BE; 
but  Z  ABO  Z  ABE. 

We  are  to  prove     A  G  >  A  E. 

Place  the  A  so  that  A  B  of  the  one  shall  coincide  with  A  B 
of  the  other. 

Draw  B  F  so  as  to  bisect  Z  EBG. 
Draw  EF. 
In  the  A  EBF&nd  GBF 

EB  =  BC,  Hyp. 

BF=BF,  Iden. 

ZEBF=Z  GBF,  Cons. 

.\  the  A  EBFtrnd  CBF&ie  equal,  §  106 

(having  two  sides  and  the  included  Z  of  one  equal  respectively  to  two  sides 
and  the  included  Z.  of  the  other). 
.'.EF=FC, 
(being  homologous  sides  of  equal  &  ). 
Now  AF+  FE>  AE,  §  96 

(the  sum  of  two  sides  of  a  A  is  greater  than  the  third  side). 
Substitute  for  FE  its  equal  FG.     Then 
AF+  FG>AE;  or, 
A  C>  A  E. 

Q.  E.  D. 


TRIANGLES.  51 


Proposition  XXXII.     Theorem. 

116.  Conversely:  If  two  sides  of  a  triangle  be  equal 
respectively  to  two  sides  of  another,  but  the  third  side  of  the 
first  triangle  be  greater  than  the  third  side  of  the  second,  then 
the  angle  opposite  the  third  side  of  the  first  triangle  is  greater 
than  the  angle  opposite  the  third  side  of  the  second. 


In  the  A  ABC  and  A' B'  C,  let  AB  =  A'B',  AC  =  A'C') 
but  BOB'  C. 

We  are  to  prove      Z  A  >  Z  A'. 
If  Z  A  =  Z  A', 

then  would       AABC  =  AA'B'C,  §  106 

(having  two  sides  and  the  included  Z.  of  the  one  equal  respectively  to  two  sides 
and  the  included  A  of  the  other), 

and  BC  =  B  C, 

(being  homologous  sides  of  equal  A ). 

And  if  A  <  A', 

then  would  BC<B'C;  §  115 

(if  two  sides  of  a  Abe  equal  respectively  to  two  sides  of  another  A,  but  the 
included  Z  of  the  first  be  greater  than  the  included  A  of  the  second,  the 
third  side  of  the  first  will  be  greater  than  the  third  side  of  the  second.) 

But  both  these  conclusions  are  contrary  to  the  hypothesis ; 

.'.  Z  A  does  not  equal  Z  A',  and  is  not  less  than  Z  A'. 

.'.ZA>Z  A'. 

Q.  E.  D 


52  GEOMETRY.  —  BOOK   I. 


Proposition  XXXIII.     Theorem. 

117.    Of  two   sides  of  a  triangle,   that  is  the  greater 
which  is  opposite  the  greater  angle. 


In  the  triangle  ABC  let  angle  AG B  be  greater  than 
angle  B. 

We  are  to  prove      A  B  >  AG. 

Draw  C  E  so  as  to  make  A  B  G  E  =  Z5. 

Then  EC  =  EB,  §114 

(being  sides  opposite  equal  A  ). 

Now  AE+EOAG,  §96 

(the  sum  of  two  sides  of  a  A  is  greater  than  the  third  side). 

Substitute  for  EG  its  e^ual  E B.     Then 

AE+  EB>  AG,  or 

A  B  >  A  G. 

Q.  E.  D. 


Ex.  ABG  and  ABB  are  two  triangles  on  the  same  base 
A  B,  and  on  the  same  side  of  it,  the  vertex  of  each  triangle 
being  without  the  other.  If  A  G  equal  A  D,  show  that  B  C 
cannot  equal  B  D. 


TRIANGLES.  53 


Proposition  XXXIV.     Theorem. 

118.   Of  two  angles  of  a  triangle,  that  is  the  greater 
which  is  opposite  the  greater  side. 

D 


C 

In  the  triangle  ABC  let  A  B  be  greater  than  A  C. 

We  are  to  prove      Z  AC  B  >  Z  /?. 

Take  A  E  equal  to  AC) 

Draw^C. 

Z  AEC  =  £  ACE,  §112 

(being  A  opposite  equal  sides). 

But  ZAEOZB,  §105 

(an  exterior  Z  of  a  A  is  greater  than  either  opposite  interior  Z  ), 

and  ZACB>ZACE. 

Substitute  for  Z  A  C E  its  equal  Z  A  EC,  then 

ZACB>ZAEC. 

Much  more  is  Z  A  CB  >  Z  B. 

Q.  E.  D. 


Ex.  If  the  angles  ABC  and  AC B,  at  the  base  of  an 
isosceles  triangle,  be  bisected  by  the  straight  lines  B .0,  CD, 
show  that  BBC  will  be  an  isosceles  triangle. 


54  GEOMETRY. BOOK   I. 


Proposition  XXXV.     Theorem. 
119.  The  three  bisectors  of  the  three  angles  of  a  triangle 
meet  in  a  point. 


Let  the  two  bisectors  of  the   angles  A  and  C  meet 
at  0,  and  0  B  be  drawn. 

We  are  to  prove      B  0  bisects  the  Z  B. 

Draw  the  Jfc  OK,  OP,  and  OH. 

Inthert.  A  0 C K and  OOP, 

OC=OC,  Iden. 

Z00K  =  Z00P,  Cons. 

.-.A  OCK=A  OOP,  §  110 

(having  the  hypotenuse  and  an  acute  Z  of  the  one  equal  respectively  to  the 
hypotenuse  and  an  acute  Z  of  tlie  other). 

.'.  OP=OK, 

(homologous  sides  of  equal  & ). 

In  the  rt.  A  OA  P  and  OAH, 

OA  =  OA,  Iden. 

ZOAP  =  ZOAH,  Cons. 

.'.AOAP  =  AOAH,  §110 

{having  the  hypotenuse  and  an  acute  Z  of  the  one  equal  respectively  to  the 
hypotenuse  and  an  acute  Z  of  the  other). 

.'.OP=OH, 

(being  homologous  sides  of  equal  &  ). 
But  we  have  already  shown  0  P  *=  0  K, 

.\OII=  OK,  Ax.  1 

Now  in  rt.  A  0  HB  and  0  KB 


TRIANGLES.  55 


OH  =  OK,  and  0  B  =  0  B, 

.'.AOHB  =  A  OKB,  §  109 

{having  the  hypotenuse  and  a  side  of  the  one  equal  respectively  to  the  hypote- 
nuse and  a  side  of  tJie  other). 


../.  OBH  =  Z  OBKt 

(being  homologous  A  of  equal  &. ). 


Q.  E.  D. 


Proposition  XXXVI.     Theorem. 
120.    The   three  perpendiculars  erected  at  the  middle 
joints  of  the  three  sides  of  a  triangle  meet  in  a  point. 
A 


F 

Let  DD',  EE',  FF',  be  three  perpendiculars  erected 
at  D,  E,  F,  the  middle  points  of  A  B,  A  C,  and  B  C. 

We  are  to  prove  they  meet  in  some  point,  as  0. 

The  two  Ji  D  D'  and  E  E'  meet,  otherwise  they  would  be 
parallel,  and  A  B  and  A  C,  being  _l§  to  these  lines  from  the  same 
point  A,  would  be  in  the  same  straight  line; 

but  this  is  impossible,  since  they  are  sides  of  a  A. 

Let  0  be  the  point  at  which  they  meet. 

Then,  since  0  is  in  D  L1,  which  is  _L  to  A  B  at  its  middle 
point,  it  is  equally  distant  from  A  and  B.  §  59 

Also,  since  0  is  in  E  E',  _L  to  A  C  at  its  middle  point,  it  is 
equally  distant  from  A  and  C. 

.'.  0  is  equally  distant  from  B  and  C ; 

.'.  0  is  in  FF  _L  to  B  C  at  its  middle  point,  §  59 

(the  locus  of  all  points  equally  distant  from  the  extremities  of  a  straight  line 

is  the  ±  erected  at  the  middle  of  that  line). 

Q.  E.  D. 


56  GEOMETRY. BOOK   I. 

Proposition  XXXVII.     Theorem. 

121.   The  three  perpendiculars  from  the  vertices  of  a  tri- 
angle to  the  opposite  sides  meet  in  a  point. 


In  the  triangle  ABC,  let  B P,  AH,  G  K,  be  the  per- 
pendiculars from  the  vertices  to  the  opposite 
sides. 

We  are  to  prove  they  meet  in  some  point,  as  0. 
Through  the  vertices  A,  B,  C,  draw 

A'B'  II  to  BO, 
A'  C  II  to  A  C, 
B'  C  II  to  A  B. 
In  the  A  A B A1  and  ABC,  we  have 

AB  =  AB,  Iden. 

ZABA'  =  ZBAC,  §68 

(being  alternate  interior  A  ), 

ZBAA'  =  ZABC.  §68 

.-.  A  ABA'  =  A  ABC,  §  107 

{having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other). 

.'.A'B  =  AC, 

(being  homologous  sides  of  egual  A  ). 


TRIANGLES.  57 


In  the  A  C  B  C  and  A  B  C, 

BC  =  BC,  Iden. 

ZCBC'  =  Z.BCAt  §68 

{being  alternate  interior  A ). 

ZBCC'  =  ZOBA.  §68 

.\ACBC'  =  AABC,  §  107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other). 

.\BG'  =  AC1 

(being  homologous  sides  of  equal  A  ). 

But  we  have  already  shown  A1  B  =  A  0, 

.\A'B  =  BC,  Ax.  1. 

.\  B  is  the  middle  point  of  A'  C 

Since  B  P  is  -L  to  A  C,  Hyp. 

it  is  J_  to  A'  C,  §  67 

(a  straight  line  which  is  ±  to  one  of  two  lis  is  _L  to  the  other  also). 

But  B  is  the  middle  point  of  A'  C ; 

.'.  BP  is  J_  to  A'  C  at  its  middle  point. 

In  like  manner  we  may  prove  that 

A  H  is  J_  to  A'  B'  at  its  middle  point, 

and  C  K  _L  to  B'  C  at  its  middle  point. 

.'.  B  P,  A  H,  and  C K  are  Js  erected  at  the  middle  points 
of  the  sides  of  the  A  A'  B'  C 

.'.  these  J§  meet  in  a  point.  §  120 

(the  three  J§  erected  at  the  middle  points  of  the  sides  of  a  A  meet  in  a  point). 

Q.  E.  D. 


58 


GEOMETRY. BOOK  I. 


On  Quadrilaterals. 

122.  Def.    A  Quadrilateral  is  a  plane  figure  bounded  by- 
four  straight  lines. 

123.  Def.    A  Trapezium  is  a  quadrilateral  which  has  no 
two  sides  parallel. 

124.  Def.    A  Trapezoid  is  a  quadrilateral  which  has  two 
sides,  and  only  two  sides,  parallel. 

125.  Def.    A  Parallelogram  is  a  quadrilateral  which  has 
its  opposite  sides  parallel. 


TRAPEZIUM. 


PARALLELOGRAM. 


126.  Def.    A  Rectangle  is  a  parallelogram  which  has  its 
angles  right  angles. 

127.  Def.    A   Square  is   a   parallelogram   which   has  its 
angles  right  angles,  and  its  sides  equal. 

128.  Def.    A  Rhombus  is  a  parallelogram  which  has  its 
sides  equal,  but  its  angles  oblique  angles. 

129.  Def.    A  Rhomboid  is  a  parallelogram  which  has  its 
angles  oblique  angles. 

The  figure  marked  parallelogram  is  also  a  rhomboid. 


RECTANGLE. 


QUADRILATERALS.  59 


130.  Def.  The  side  upon  which  a  parallelogram  stands, 
and  the  opposite  side,  are  called  its  lower  and  upper  bases ;  and 
the  parallel  sides  of  a  trapezoid  are  called  its  bases. 

131.  Def.  The  Altitude  of  a  parallelogram  or  trapezoid  is 
the  perpendicular  distance  between  its  bases. 

132.  Def.  The  Diagonal  of  a 
quadrilateral  is  a  straight  line  joining 
any  two  opposite  vertices. 

Proposition  XXXVIII.     Theorem. 

133.  The  diagonal  of  a  parallelogram  divides  the  figure 
into  two  equal  triangles. 


A  E 

Let  ABC E  be  a  parallelogram,  and  A  C  its  diagonal. 
We  are  to  prove      AABC=AA  EC. 
In  the  A  ABC  and  A  EC 

AC  =  AC,  Iden. 

ZACB  =  ZCAE,  §68 

( being  alt.  -int.  A ). 
ZCAB  =  ZACE,  §68 

.\AABC  =  AAEC,  §  107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  tiro 

adj.  A  of  the  other). 

Q.  E.  D. 


60  GEOMETRY. BOOK   I. 


Proposition  XXXIX.     Theorem. 

134.  In  a  parallelogram  the  opposite  sides  are  equal, 
and  the  opposite  angles  are  equal. 

B  G 


A  E 

Let  the  figure  ABC E  be  a  parallelogram. 

We  are  to  prove    B  C  =  A  E,  and  AB  =  EC, 

also,     ZB  =  ZE,andZBAE  =  ZBCE. 

Draw  A  C. 

AABC  =  AAEC,  §133 

(the  diagonal  of a  O  divides  the  figure  into  two  equal  &  ). 

.-.BC  =  AE1 

and  AB  =  CE, 

(being  homologous  sides  of  equal  A  ). 

ZB  =  Z  E, 

(being  homologous  A  of  equal  A ). 

Z  BAC  =  Z  ACEf 

and  ZEAC  =  ZACB, 

{being  homologous  A  of  equal  A). 

Add  these  last  two  equalities,  and  we  have 

ZBAC  +  Z  EAC  =  ZACE+ZACB; 
or,  ZBAE  =  ZBCE. 

Q.  E.  D. 

1 35.  Corollary.  Parallel  lines  comprehended  between  par- 
allel lines  are  equal. 


QUADRILATERALS.  61 


Proposition  XL.     Theorem. 

136.  If  a  quadrilateral  have  two  sides  equal  and  par- 
allel, then  the  other  two  sides  are  equal  and  parallel,  and  the 
figure  is  a  parallelogram. 

B  C 


Let  the  figure  ABC  E  be  a  quadrilateral,  having  the 
side  A  E  equal  and  parallel  to  B  C. 

We  are  to  prove     A  B  equal  and  II  to  E  C. 

Draw  A  G. 
In  the  A  A  B  C  and  A  E  C 

BC  =  AE,  Hyp. 

AC  =  AC,  Iden. 

ZBCA=ZCAE,  §68 

(being alt. -int.  A). 

.\AABC  =  AACE,  §106 

(kceoing  two  sides  and  the  included  Z.  of  the  one  equal  respectively  to  two  sides 
and  the  included  Z.  of  the  other). 

.'.AB  =  EC, 

(being  homologous  sides  of  equal  A  ). 

Also,  ZBAC  =  ZACE, 

(being  homologous  A  of  equal  A  )  ; 

.'.A  Bis  II  to  EG,  §69 

(when  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  alt.  -int.  A  be 
equal  the  lines  are  parallel). 

.*.  the  figure  ABC  E  is  a  O,  §  125 

(the  opposite  sides  being  parallel). 

Q.  E.  D. 


62  GEOMETRY.  —  BOOK   I. 

Proposition  XLL     Theorem. 

137.  If  in  a  quadrilateral  the  opposite  sides  be  equal,  the 
figure  is  a  parallelogram. 

B  C 


A 

Let    the  figure    A  B  G  E    be   a    quadrilateral   having 
BG  =  AE  and  AB  =  EC. 

We  are  to  prove  figure     ABC  E  a  E3. 

Draw  A  C. 

In  the  A  ABC  &nd  A  EG 

BG  =  AEf  Hyp. 

AB  =  GE,  Hyp. 

A  C  =  A  G,  Iden. 

.\AABG  =  AAEC,  §  108 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

.'.ZACB  =  Z  GAE, 

and  ZBAC  =  ZAGE, 

(being  homologous  A  of  equal  A  ). 

.'.BC  is  II  toAE, 

and  A  Bis  II  to  EC,  §69 

(when  two  straight  lines  lying  in  the  same  plane  are  cut  by  a  third  straight 
line,  if  the  alt.-int.  A  be  equal,  the  lines  are  parallel). 

.'.  the  figure  A  B  G  E  is  a  O,  §  125 

(having  its  opposite  sides  parallel). 


QUADRILATERALS. 


63 


Proposition  XLII.     Theorem. 
138.  The  diagonals  of  a  parallelogram  bisect  each  other. 
B 


Let  the  figure  ABCE  be  a  parallelogram,  and  let 
the  diagonals  A  C  and  BE  cut  each  other  at  0. 

We  are  to  prove      AO  =  OC,  and  B  0  =  0  E. 

In  the  A  A  0  E  and  B  0  C 


AE=BC, 

§  134 

(being  opposite  sides  of  a  CD '), 

Z  OAE  =  Z  OCB, 

§  68 

(being  alt.  -int.  A  ), 

Z  OEA=Z  OBC; 

§  68 

.'.AAOE  =  ABOC, 

§  107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other). 

.'.AO  =  OC, 

and  BO  =  0  E. 

(being  homologous  sides  of  equal  A  ). 

Q.  E.  D. 


64  GEOMETRY.  —  BOOK  I. 


Proposition  XLIII.     Theorem. 

139.   The  diagonals  of  a  rhombus  bisect  each  other  at 

right  angles. 

A  E 


B  C 

Let   the  figure    A  B  C  E    be   a   rhombus,   having  the 
diagonals  AC  and  BE  bisecting  each  other  at  0. 

We  are  to  prove      Z  A  0  E  and  Z  A  0  B  rt.  A. 

In  the  A  A  0  E  and  A  0  B, 

AE  =  AB,  §128 

(being  sides  of  a  rhombus)  ; 

OE=OB,  §138 

(the  diagonals  of  a  EJ  bisect  each  other)  ; 

AO  =  AO,  Iden. 

.'.AAOE=AAOB,  §  108 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other) ; 

.'.ZAOE  =  ZAOB, 

(being  homologous  A  of  equal  A  )  ; 
.'.  Z  A  0  E  and  Z  A  0  B  are  rt.  A.  §  25 

(  When  one  straight  line  meets  another  straight  line  so  as  to  make  the  adj.  A 
I,  each  Z.  is  art.  Z). 

Q.  E.  D, 


QUADRILATERALS.  65 


Proposition  XLIV.     Theorem. 

140.  Two  parallelograms,  having  two  sides  and  the  in- 
cluded angle  of  the  one  equal  respectively  to  Uoo  sides  and  the 
included  angle  of  the  other,  are  equal  in  all  respects. 

B  C  B  a 


A' 


In   the  parallelograms    A  B  C  I)    and  A'  B'  C  D't    le 
AB  =  A'B'9  AD  =  A'D't  and  A  A  =  A  A'. 

We  are  to  prove  that  the  UD  are  equal. 

Apply  O  A  BCD  to  O  A' B' CD',  so  that  A  D  will  fall 
on  and  coincide  with  A'  L'. 

Then  A  B  will  fall  on  A' B', 
{for  ZA  =  ZA',bU  hi/p.), 

and  the  point  B  will  fall  on  B'y 
{for  AB=  A'  B\  by  ifjp.). 

Now,  B  C  and  B'  C  are  "both  II  to  A'  D'  and  are  drawn 

through  point  B' '; 

.'.  the  lines  B  C  and  B'  C  coincide,  §  66 

and  C  falls  on  B'  C  or  B'  C  produced. 

In  like  manner  D  C  and  D'  C  are  II  to  A'  B'  and  are  drawn 
through  the  point  D'. 

.'.  D  C  and  D'  C  coincide ;  §  66 

.*.  the  point  C  falls  on  D'  C,  or  B'  C  produced  ; 

.'.  C  falls  on  both  B'  C  and  D'  O '; 

.'.  C  must  fall  on  a  point  common  to  both,  namely,  C. 

.'.the  two  UJ  coincide,  and  are  equal  in  all  respects. 

Q.  E.  D. 

141.  Corollary.    Two  rectangles  having  the  same  base  and 

altitude  are  equal ;  for  they  may  be  applied  to  each  other  and 

will  coincide. 


66  GEOMETRY. BOOK   I. 


Proposition  XLV.     Theorem. 

142.  The  straight  line  which  connects  the  middle  points 
of  the  non-parallel  sides  of  a  trapezoid  is  parallel  to  the  par- 
allel sides j  and  is  equal  to  half  their  sum. 

A  F      E 


Let  SO  be  the  straight  line  joining  the  middle  points 
of  the  non-parallel  sides  of  the  trapezoid  ABCE. 

We  are  to  prove      SO  II  to  A  E  and  B  0 ; 
also      SO  =  ^(AE  +  BC). 

Through  the  point  0  draw  FH  II  to  A  B, 

and  produce  B  0  to  meet  FO H at  H. 

In  the  A  FOE  and  COH 

OE=00,  Cons. 

ZOEF=ZOCH,  §68 

(being  alt.  -int.  A  ), 

ZFOE  =  ZOOfft  §49 

(being  vertical  A  ). 

.\AFOE  =  ACOH,  §107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other). 


QUADRILATERALS.  67 


.'.FE=C  H, 

and  OF=OH, 

(being  homologous  sides  of  equal  A ). 

Now  FH  =  AB,  §  135 

( II  lines  comprehended  between  II  lines  are  equal)  ; 

.'.FO  =  AS.  Ax.  7. 

.'.  the  figure  AFOSis&CJ,  §  136 

(having  two  opposite  sides  equal  and  parallel). 

.'.SO  is  II  to  AF,  §  125 

(being  opposite  sides  of  a  Of). 

£0  is  also   II  to  BC, 
(a  straight  line  II  to  one  of  two  II  lines  is  II  to  the  other  also). 


Now 

SO  =  AF, 

§125 

(being  opposite  sides  of  a  CJ)t 

and 

SO  =  BH. 

§125 

But 

AF=AE-FEf 

and 

BH=BC+  CH. 

Substitute  for  A  F  and  BH  their  equals,  A  E  —  FE  and 
BC+  CH, 

and  add,  observing  that  CH=  FE; 

then  2SO  =  AE+BC. 

.-.SO  =  b(AE+  BC). 

Q.  E.  D. 


68 


GEOMETRY. 


BOOK   I. 


On  Polygons  in  General. 

143.  Def.  A  Polygon  is  a  plane  figure  bounded  by  straight 
lines. 

144.  Def.  The  bounding  lines  are  the  sides  of  the  polygon, 
and  their  sum,  as  A  B  +  B  C  +  CD,  etc.,  is  the  Perimeter  of 
the  polygon. 

The  angles  which  the  adjacent  sides  make  with  each  other 
are  the  angles  of  the  polygon. 

145.  Def.  A  Diagonal  of  a  polygon  is  a  line  joining  the 
vertices  of  two  angles  not  adjacent. 

B  B< 


C      A' 


D      F' 


e  m 

146.  Def.  An  Equilateral  polygon  is  one  which  has  all  its 
sides  equal. 

147.  Def.  An  Equiangular  polygon  is  one  which  has  all 
its  angles  equal. 

148.  Def.  A  Convex  polygon  is  one  of  which  no  side, 
when  produced,  will  enter  the  surface  bounded  by  the  perimeter. 

149.  Def.  Each  angle  of  such  a  polygon  is  called  a  Salient 
angle,  and  is  less  than  two  right  angles. 

150.  Def.  A  Concave  polygon  is  one  of  which  two  or  more 
sides,  when  produced,  will  enter  the  surface  bounded  by  the 
perimeter. 

151.  Def.    The  angle  ED  E  is  called  a  Re-entrant  angle. 
When  the  term  polygon  is  used,  a  convex  polygon  is  meant. 
The  number  of  sides  of  a  polygon  is  evidently  equal  to  the 

number  of  its  angles. 

By  drawing  diagonals  from  any  vertex  of  a  polygon,  the  fig- 
ure may  be  divided  into  as  many  triangles  as  it  has  sides  less  two. 


POLYGONS.  69 


152.  Def.  Two  polygons  are  Equal,  when  they  can  be 
divided  by  diagonals  into  the  same  number  of  triangles,  equal 
each  to  each,  and  similarly  placed;  for  the  polygons  can  be 
applied  to  each  other,  and  the  corresponding  triangles  will  evi- 
dently coincide.  Therefore  the  polygons  will  coincide,  and  be 
equal  in  all  respects. 

153.  Def.  Two  polygons  are  Mutually  Equiangular,  if  the 
angles  of  the  one  be  equal  to  the  angles  of  the  other,  each  to 
each,  when  taken  in  the  same  order;  as  the  polygons  ABC DE F, 
and  A1  B<  C  D'  E>  F,  in  which  Z  A  =  Z  A',  Z  B  =  Z  B', 
ZC  =  ZC,  etc. 

154.  Def.  The  equal  angles  in  mutually  equiangular  poly- 
gons are  called  Homologous  angles ;  and  the  sides  which  lie 
between  equal  angles  are  called  Homologous  sides. 

155.  Def.  Two  polygons  are  Mutually  Equilateral,  if  the 
sides  of  the  one  be  equal  to  the  sides  of  the  other,  each  to  each, 
when  taken  in  the  same  order. 


Fig.  1.  Fig.  2.  Fig.  3. 

Two  polygons  may  be  mutually  equiangular  without  being 
mutually  equilateral ;  as  Figs.  1  and  2. 

And,  except  in  the  case  of  triangles,  two  polygons  may  be 
mutually  equilateral  without  being  mutually  equiangular;  as 
Figs.  3  and  4. 

If  two  polygons  be  mutually  equilateral  and  equiangular, 
they  are  equal,  for  they  may  be  applied  the  one  to  the  other  so 
as  to  coincide. 

156.  Def.  A  polygon  of  three  sides  is  a  Trigon  or  Tri- 
angle;  one  of  four  sides  is  a  Tetragon  or  Quadrilateral ;  one  of 
five  sides  is  a  Pentagon ;  one  of  six  sides  is  a  Hexagon ;  one  of 
seven  sides  is  a  Heptagon ;  one  of  eight  sides  is  an  Octagon  ;  one 
of  ten  sides  is  a  Decagon  ;  one  of  twelve  sides  is  a  Dodecagon. 


70 


GEOMETRY. BOOK   I. 


Proposition  XL VI.     Theorem. 

157.  The  sum  of  the  interior  angles  of  a  polygon  is 
equal  to  two  right  angles,  taken  as  many  times  less  two  as 
the  figure  has  sides. 


A  B 

Let  the  figure  A  BO  DBF  be  a  polygon  having  n  sides. 

We  are  to  prove 

ZA  +  ZB+ZC,  etc.,  =  2  rt.  A  (n  —  2). 
From  the  vertex  A  draw  the  diagonals  AC,  A  D,  and  A  E. 

The  sum  of  the  A  of  the  A  =  the  sum  of  the  angles  of  the 
polygon. 

Now  there  are  (n  —  2)  A, 


and  the  sum  of  the  A  of  each  A  =  2  rt.  A. 


98 


/.the  sum  of  the  A  of  the  A,  that  is,  the  sum  of  the  A  of 
the  polygon  =  2  rt.  A  (?i  —  2). 

Q.  E.  D. 

158.  Corollary.  The  sum  of  the  angles  of  a  quadrilateral 
equals  two  right  angles  taken  (4  —  2)  times,  i.  e.  equals  4  right 
angles ;  and  if  the  angles  be  all  equal,  each  angle  is  a  right 
angle.     In  general,  each  angle  of  an  equiangular  polygon  of  n 

sides  is  equal  to  — I- 1  right  angles. 


POLYGONS.  71 


Proposition  XLVII.     Theorem. 

159.  The  exterior  angles  of  a  polygon,  made  by  produ- 
cing each  of  its  sides  in  succession,  are  together  equal  to  four 
right  angles. 


Let  the  figure  ABCDE  be  a  polygon,  having  its  sides 
produced  in  succession. 

We  are  to  prove  the  sum  of  the  ext.  A  =  4  rt.  A. 

Denote  the  int.  A  of  the  polygon  by  A,  B,C,D,  E ; 

and  the  ext.  A  by  a,  b,  c,  d,  e. 

ZA  +  Za=2vtA,  §34 

(being  siq).  -adj.  A  ). 

Z  B  +  A  b  =  2  rt.  A.  §  34 

In  like  manner  each  pair  of  adj.  A  =  2  rt.  A ; 

.'.the  sum  of  the  interior  and  exterior  A  =  2  rt.  A  taken 
as  many  times  as  the  figure  has  sides, 

or,  2  n  rt.  A. 

But  the  interior  A  =  2  rt.  A  taken  as  many  times  as  the 
figure  has  sides  less  two,  =  2  rt.  A  (n  —  2), 

or,  2  n  rt.  A  —  4  rt.  A. 

.'.  the  exterior  A  =  4  rt.  A. 

Q.  E.  D. 


72  GEOMETRY. BOOK   I. 


Exercises. 

1.  Show  that  the  sum  of  the  interior  angles  of  a  hexagon  is 
equal  to  eight  right  angles. 

2.  Show  that  each  angle  of  an  equiangular  pentagon  is  f  of 
a  right  angle. 

3.  How  many  sides  has  an  equiangular  polygon,  four  of 
whose  angles  are  together  equal  to  seven  right  angles? 

4.  How  many  sides  has  the  polygon  the  sum  of  whose  in- 
terior angles  is  equal  to  the  sum  of  its  exterior  angles  1 

5.  How  many  sides  has  the  polygon  the  sum  of  whose  in- 
terior angles  is  double  that  of  its  exterior  angles  1 

6.  How  many  sides  has  the  polygon  the  sum  of  whose 
exterior  angles  is  double  that  of  its  interior  angles  1 

7.  Every  point  in  the  bisector  of  an  angle  is  equally  distant 
from  the  sides  of  the  angle ;  and  every  point  not  in  the  bisector, 
but  within  the  angle,  is  unequally  distant  from  the  sides  of  the 
angle. 

8.  B  A  C  is  a  triangle  having  the  angle  B  double  the  angle 
A.  If  B  D  bisect  the  angle  B,  and  meet  A  C  in  D,  show  that 
B  D  is  equal  to  A  D. 

9.  If  a  straight  line  drawn  parallel  to  the  base  of  a  triangle 
bisect  one  of  the  sides,  show  that  it  bisects  the  other  also ;  and 
that  the  portion  of  it  intercepted  between  the  two  sides  is  equal 
to  one  half  the  base. 

10.  ABC D  is  a  parallelogram,  E  and  F  the  middle  points 
of  A  D  and  B  C  respectively  j  show  that  B  E  and  D  F  will 
trisect  the  diagonal  A  C. 

11.  If  from  any  point  in  the  base  of  an  isosceles  triangle 
parallels  to  the  equal  sides  be  drawn,  show  that  a  parallelogram 
is  formed  whose  perimeter  is  equal  to  the  sum  of  the  equal 
sides  of  the  triangle. 

12.  If  from  the  diagonal  BD  of  a  square  AB  CD,  BE  be 
cut  off  equal  to  B  C,  and  E  F  be  drawn  perpendicular  to  B  D, 
show  that  D  E  is  equal  to  E  F,  and  also  to  F  C. 

13.  Show  that  the  three  lines  drawn  from  the  vertices  of  a 
triangle  to  the  middle  points  of  the  opposite  sides  meet  in  a 
point. 


BOOK  II. 

CIRCLES. 


Definitions. 

160.  Def.  A  Circle  is  a  plane  figure  bounded  by  a  curved 
line,  all  the  points  of  which  are  equally  distant  from  a  point 
within  called  the  Centre. 

161.  Def.  The  Circumference  of  a  circle  is  the  line  which 
bounds  the  circle. 

162.  Def.  A  Radius  of  a  circle  is  any  straight  line  drawn 
from  the  centre  to  the  circumference,  as  0  A,  Fig.  1. 

163.  Def.  A  Diameter  of  a  circle  is  any  straight  line  pass- 
ing through  the  centre  and  having  its  extremities  in  the  circum- 
ference, as  A  B,  Fig.  2. 

By  the  definition  of  a  circle,  all  its  radii  are  equal.  Hence, 
all  its  diameters  are  equal,  since  the  diameter  is  equal  to  twice 
the  radius. 

M  M 


Fig.  1. 


164.  Def.  An  Arc  of  a  circle  is  any  portion  of  the  circum- 
ference, as  A  M  B,  Fig.  3. 

165.  Def.  A  Semi-circumference  is  an  arc  equal  to  one 
half  the  circumference,  as  A  M  B,  Fig.  2. 

166.  Def.  A  Chord  of  a  circle  is  any  straight  line  having 
its  extremities  in  the  circumference,  as  A  B,  Fig.  3. 

Every  chord  subtends  two  arcs  whose  sum  is  the  cir- 
cumference. Thus  the  chord  A  B,  (Fig.  3),  subtends  the  arc 
A  MB  and  the  arc  A  D  B.  Whenever  a  chord  and  its  arc  are 
spoken  of,  the  less  arc  is  meant  unless  it  be  otherwise  stated. 


74  GEOMETRY.  —  BOOK   II. 

167.  Def.  A  Segment  of  a  circle  is  a  portion  of  a  circle 
enclosed  by  an  arc  and  its  chord,  as  A  M  B,  Fig.  1. 

168.  Def.  A  Semicircle  is  a  segment  equal  to  one  half  the 
circle,  as  A  D  C,  Fig.  1. 

169.  Def.  A  Sector  of  a  circle  is  a  portion  of  the  circle 
enclosed  by  two  radii  and  the  arc  which  they  intercept,  as  A  C  P 
Fig.  2. 

170.  Def.  A  Tangent  is  a  straight  line  which  touches  the 
circumference  but  does  not  intersect  it,  however  far  produced. 
The  point  in  which  the  tangent  touches  the  circumference  is 
called  the  Point  of  Contact,  or  Point  of  Tangency. 

171.  Def.  Two  Circumferences  are  tangent  to  each  other 
when  they  are  tangent  to  a  straight  line  at  the  same  point. 

172.  Def.  A  Secant  is  a  straight  line  which  intersects  the 
circumference  in  two  points,  as  A  D,  Fig.  3. 


173.  Def.    A  straight  line  is  Inscribed  in  a  circle  when  its 
extremities  lie  in  the  circumference  of  the  circle,  as  A  B,  Fig.  1. 

An  angle  is  inscribed  in  a  circle  when  its  vertex  is  in  the 
circumference  and  its  sides  are  chords  of  that  circumference,  as 
ZABdTig.  1. 

A  polygon  is  inscribed  in  a  circle  when  its  sides  are  chords 
of  the  circle,  as  A  A  B  C,  Fig.  1. 

A  circle  is  inscribed  in  a  polygon  when  the  circumference 
touches  the  sides  of  the  polygon  but  does  not  intersect  them, 
as  in  Fig.  4. 

174.  Def.    A  polygon  is  Circumscribed  about  a  circle  when 
all  the  sides  of  the  polygon  are  tangents  to  the  circle,  as  in  Fig.  4. 

A  circle  is  circumscribed  about  a  polygon  when  the  circumfer- 
ence passes  through  all  the  vertices  of  the  polygon,  as  in  Fig.  1. 


STRAIGHT   LINES   AND   CIRCLES. 


75 


175.  Dep.  Equal  circles  are  circles  which  have  equal  radii 
For  if  one  circle  be  applied  to  the  other  so  that  their  centres 
coincide  their  circumferences  will  coincide,  since  all  the  points 
of  both  are  at  the  same  distance  from  the  centre. 

176.  Every  diameter  bisects  the  circle 
and  its  circumference.  For  if  we  fold  over 
the  segment  A  MB  on  A  B  as  an  axis  until 
it  comes  into  the  plane  of  A  P  B,  the  arc 
A  MB  will  coincide  with  the  arc  AP  B\ 
because  every  point  in  each  is  equally  dis- 
tant from  the  centre  0. 


Proposition  I.     Theorem. 

177.  The  diameter  of  a  circle  is  greater  than  any  other 
chord. 

Let  A  B  be  the  diameter  of  the  circle 
A  MB,  and  A  E  any  other  chord. 

We  are  to  prove    A  B  >  A  E. 

From  C,  the  centre  of  the  O,  draw  C  E. 
CE=CB, 

(being  radii  of  the  same  circle). 

But  AC+CE>AE, 


§96 


{the  sum  of  two  sides  ofaA>  tlie  third  side). 
Substitute  for  0  E,  in  the  above  inequality,  its  equal  CB. 
Then  A  C  +  CB  >  A  E,  or 

AB>  AE.      ' 

Q.  E.  D. 


76  GEOMETRY.  —  BOOK   II. 

Proposition  II.     Theorem. 

178.  A  straight  line  cannot  intersect  the  circumference 
of  a  circle  in  more  than  two  points. 

M 


P 

Let  HK  be  any  line  cutting  the  circumference  A  MP. 

We  are  to  prove  that  HK  can  intersect  the  circumference 
in  only  two  points. 

If  it  be  possible,  let  HK  intersect  the  circumference  in  three 
points,  H,  P,  and  K. 

From  0,  the  centre  of  the  O,  draw  the  radii  OH,  OP, 
and  OK. 

Then  OH,  OP,  and  0  K  are  equal,  §  1 63 

(being  radii  of  the  same  circle). 

.'.if  HK  could  intersect  the  circumference  in  three  points, 
we  should  have  three  equal  straight  lines  OH,  OP,  and  0 K 
drawn  from  the  same  point  to  a  given  straight  line,  which  is 
impossible,  §  56 

(only  two  equal  straight  lines  can  be  drawn  from  a  point  to  a  straight  line). 

.*.  a  straight  line  can  intersect  the  circumference  in  only 
two  points. 

Q.  E.  D. 


STRAIGHT    LINES    AND    CIRCLES.  77 


Proposition  III.     Theorem. 

179.  In  the  same  circle,  or  equal  circles,  equal  angles 
at  the  centre  intercept  equal  arcs  on  the  circumference. 


P  F 

In  the  equal  circles  ABP  and  A'B'P'  let  ZO=ZO'. 

We  are  to  prove      arc  R  S  =  arc  R'  S'. 

Apply  Oi^PtoO  A'B'P1, 

so  that  Z  0  shall  coincide  with  Z.  0'. 

The  point  R  will  fall  upon  7?',  §  176 

(for  0  R=  Of  R',  being  radii  of  equal  (D), 

and  the  point  £  will  fall  upon  S',  §  176 

(for  0S=  Of  S',  being  radii  of  equal  (D). 

Then  the  arc  R  S  must  coincide  with  the  arc  R'S'. 
For,  otherwise,  there  would  be  some  points  in  the  circumference 
unequally   distant  from  the  centre,  which   is  contrary  to   the 
definition  of  a  circle.  §  160 

Q.  E.  D. 


78  GEOMETRY. BOOK   II. 


Proposition  IV.     Theorem. 

180.  Conversely  :  In  the  same  circle,  or  equal  circles, 
equal  arcs  subtend  equal  angles  at  the  centre. 


In  the  equal   circles  ABP    and  A' B' P'  let  arc   MS 
=  arc  R'S'. 

We  are  to  prove    A  ROS=Z  R' 0' S'. 

Apply  Q)  ABP  to  Q  A' B i», 

so  that  the  radius  0  R  shall  fall  upon  0'  R'. 

Then  S,  the  extremity  of  arc  PS, 

will  fall  upon  S',  the  extremity  of  arc  R'  Sf, 
(for  RS=R>S'ibyKyp.). 

.'.  0  S  will  coincide  with  0'  S',  §  18 

(their  extremities  being  the  same  points). 

.'.  Z.  RO S  will  coincide  with,  and  be  equal  to,  Z.  R'  0' 8'. 

Q.  E.  D. 


STRAIGHT   LINES   AND   CIRCLES. 


79 


Proposition  V.     Theorem. 

181.  In  the  same  circle,  or  equal  circles,  equal  arcs  are 
subtended  by  equal  chords. 


In   the    equal   circles  ABP   and  A' B' P'  let   arc  RS 
=  arc  R'S'. 

We  are  to  prove      chord  R  S =  chord  R'  S'. 

Draw  the  radii  0  R,  0  S,  0'  R',  and  0'  S'. 

In  the  A  R  OS  and  R'  0' S' 


OR=0'R', 

(being  radii  of  equal  ©), 

OS=0'S', 

ZO  =  ZO', 

(equal  arcs  in  equal  ©  subtend  equal  A  at  the  centre). 


§176 

§176 
§  180 

§  106 


.'.A  ROS  =  A  R'O'S', 

(two  sides  and  the  included  Z  of  the  one  being  equal  respectively  to  two  sides 

and  the  included  A  of  the  other). 

.'.  chord  RS  =  chord  R'S', 
(being  homologous  sides  of  equal  A  ). 

Q.  E.  D. 


80  GEOMETRY. BOOK   II. 


Proposition  VI.     Theorem. 

182.  Conversely  :   In  the  same  circle,  or  equal  circles, 
equal  chords  subtend  equal  arcs. 


In  the  equal  circles  ABP  and  A' B' P',  let   chord  RS 
=  chord  R'S'. 

We  are  to  prove      arc  R  S  =  arc  R'  S'. 

Draw  the  radii  0  R,  0  S,  0' R',  and  0' S'. 

In  the  AROSsui&R'O'S' 

RS=R'S',  Hyp. 

OR  =  0'R'}  §176 

(being  radii  of  equal  CD), 

OS=0'S';  §176 

.\&ROS  =  AR'0'S',  §108 

(three  sides  of  the  one  being  equal  to  three  sides  of  the  other). 

.\Z  0=Z  0', 

(being  homologous  A  of  equal  &). 

.'.arc  RS  =  &rc  R'S',  §  179 

(in  the  same  O,  or  equal  (D,  equal  A  at  the  centre  intercept  equal  arcs  on  the 

circumference). 

Q.  E.  D. 


STRAIGHT    LINES    AND    CIRCLES. 


81 


Proposition  VII.     Theorem. 

183.   The  radius  perpendicular  to  a   chord   bisect*   the 
chord  and  the  arc  subtended  bj/  it. 


Let  AB  be  the  chord,  and  let  the  radius  C  S  be  per- 
pendicular to  A  B  at  the  point  if. 

We  are  to  prove      A  M  =  BM,  and  arc  A  S  =  arc  B  S. 

Draw  CA  ami  C  B. 


CA  =  CB, 

(being  radii  of  tlie  same  O)  ; 


§  84 


.'.A  AC  B  is  isosceles, 
(the  opposite  sides  being  equal)  ; 

.*.  k-  08  bisects  the  base  A  B  and  the  Z.  C,         §  113 

(the  ±  drawn  from  the  vertex  to  the  base  of  an  isosceles  A  bisects  tlie  base  and 
the  Z  at  the  vertex). 

.'.AM=BM. 

Also,  since  ZACS  =  ZBGS, 

arc  A  £■—  arc  SB,  §175 

(equal  A  at  tlie  centre  intercept  equal  arcs  on  the  circumference). 

Q.  E.  D. 

184.  Corollary.  The  perpendicular  erected  at  the  middle 
of  a  chord  passes  through  the  centre  of  the  circle,  and  bisects 
the  arc  of  the  chord. 


82  GEOMETRY. BOOK    II. 


Proposition  VIII.     Theorem. 

185.  In  the  same  circle,  or  equal  circles,  equal  chords 
are  equally  distant  from  the  centre ;  and  of  two  unequal 
chords  the  less  is  at  the  greater  distance  from  the  centre. 


E  F 

In  the  circle  A  B EC  let  the  chord  A  B  equal  the  chord 
C  F,  and  the  chord  C E  be  less  than  the  chord  G F. 
Let  OP,  OH,  and  0 K  be  J§  drawn  to  these  chords 
from  the  centre  0. 

We  are  to  prove     OP  =  Oil,  and  OH<  OK. 

Join  OA  and  OC. 
In  the  rt.  A  A  OF  and  CO II 

OA=OC, 

(being  radii  of  the  same  O)  ; 

AP=CH,  §  183 

(being  halves  of  equal  chords)  ; 

.'.AAOF  =  ACOH.  §109 

.\OF=OH. 
Again,  since  CE<  CF, 

the  ZCOF<COF,  §116 

and  the  arc  CE<  the  arc  CF. 

.'.A.  OiT  will  intersect  CF  in  some  point,  as  m. 
Now  OK>Om.  Ax.  8 

But  Om>OII,  §52 

(a  _L  is  the  shortest  distance  from  a  point  to  a  straight  line). 

.'.much  more  is  OK>  OH. 

Q.  E.  D. 


STRAIGHT    LINES    AND    CIRCLES. 


83 


Proposition  IX.     Theorem. 

186.  A  straight  line  perpendicular  to  a  radius  at  its 
extremity  is  a  tangent  to  the  circle. 


Let  BA    be   the   radius,    and  MO    the   straight  line 
perpendicular  to  BA  at  A. 

We  are  to  prove      M  0  tangent  to  the  circle. 

From  B  draw  any  other  line  to  M  0,  as  B  C  H. 

BH>BA,  §52 

(a  _1_  measures  the  shortest  distance  from  a  point  to  a  straigJU  line). 

.'.  point  ^T  is  without  the  circumference. 

But  B H  is  any  other  line  than  B  A, 

.'.  every  point  of  the  line  MO  is  without  the  circumference, 
except  A. 

.'.  MO  is  a  tangent  to  the  circle  at  A.  §  171 

Q.  E.  D. 

187.  Corollary.  When  a  straight  line  is  tangent  to  a 
circle,  it  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact,  and  therefore  a  perpendicular  to  a  tangent  at  the  point 
of  contact  passes  through  the  centre  of  the  circle. 


84 


GEOMETRY. 


BOOK   II. 


Proposition  X.     Theorem. 

188.  When  two  circumferences  intersect  each  other,  the 
line  which  joins  their  centres  is  perpendicular  to  their  common 
chord  at  its  middle  point. 


Let  C  and  C  be  the  centres  of  two  circumferences 
which  intersect  at  A  and  B.  Let  A  B  be  their 
common  chord,  and  00'  join  their  centres. 

We  are  to  prove  C  C  _L  to  A  B  at  its  middle  point. 

A  _L  drawn  through  the  middle  of  the  chord  A  B  passes 
through  the  centres  C  and  C,  §  184 

(a  _L  erected  at  the  middle  of  a  chord  passes  through  the  centre  of  the  O). 
.*.  the  line  C  C,  having  two  points  in  common  with  this  J_, 
must  coincide  with  it. 

.*.  C  C  is  _L  to  A  B  at  its  middle  point. 

Q.  E.  D. 


Ex.  1.  Show  that,  of  all  straight  lines  drawn  from  a  point 
without  a  circle  to  the  circumference,  the  least  is  that  which, 
when  produced,  passes  through  the  centre. 

Ex.  2.  Show  that,  of  all  straight  lines  drawn  from  a  point 
within  or  without  a  circle  to  the  circumference,  the  greatest  is 
that  which  meets  the  circumference  after  passing  through  the 
centre. 


STRAIGHT    LINES    AND    CIRCLES. 


85 


Proposition  XL     Theorem. 

189.  When  two  circumferences  are  tangent  to  each  other 
their  point  of  contact  is  in  the  straight  line  joining  their 
centres. 


Let  the  two  circumferences,  whose  centres  are  C  and 
C,  touch  each  other  at  0,  in  the  straight  line  A  B, 
and  let  CC  be  the  straight  line  joining  their  cen- 
tres. 

We  are  to  prove      0  is  in  the  straight  line  C  C 

A  _L  to  A  B.  drawn  through  the  point  0,  passes  through  the 
centres  C  and  C,  ,  §  187 

(a  A.  to  a  tangent  at  the  point  of  contact  passes  through  tJie  centre  of  tJie  O). 

.'.  the  line  C  C,  having  two  points  in  common  with  this  _L, 
must  coincide  with  it. 


'.  0  is  in  the  straight  line  C  C. 


Q.  E.  D. 


Ex.  A  B,  a  chord  of  a  circle,  is  the  base  of  an  isosceles 
triangle  whose  vertex  C  is  without  the  circle,  and  whose  equal 
sides  meet  the  circle  in  D  and  E.  Show  that  CD  is  equal 
to  CE. 


GEOMETRY.  —  BOOK   II. 


On  Measurement. 

190.  Def.  To  measure  a  quantity  of  any  kind  is  to  find 
how  many  times  it  contains  another  known  quantity  of  the  same 
kind.  Thus,  to  measure  a  line  is  to  find  how  many  times  it  con- 
tains another  known  line,  called  the  linear  unit. 

191.  Def.  The  number  which  expresses  how  many  times 
a  quantity  contains  the  unit,  prefixed  to  the  name  of  the  unit, 
is  called  the  numerical  measure  of  that  quantity ;  as  5  yards,  etc. 

192.  Def.  Two  quantities  are  commensurable  if  there  be 
some  third  quantity  of  the  same  kind  which  is  contained  an 
exact  number  of  times  in  each.  This  third  quantity  is  called 
the  common  measure  of  these  quantities,  and  each  of  the  given 
quantities  is  called  a  multiple  of  this  common  measure. 

193.  Def.  Two  quantities  are  incommensurable  if  they 
have  no  common  measure. 

194.  Def.  The  magnitude  of  a  quantity  is  always  relative 
to  the  magnitude  of  another  quantity  of  the  same  kind.  No 
quantity  is  great  or  small  except  by  comparison.  This  relative 
magnitude  is  called  their  Ratio,  and  this  ratio  is  always  an  ab- 
stract number. 

When  two  quantities  of  the  same  kind  are  measured  by  the 
same  unit,  their  ratio  is  the  ratio  of  their  numerical  measures. 

195.  The  ratio  of  a  to  b  is  written  -,  or  a  :  b,  and  by  this 
is  meant :  * 

How  many  times  b  is  contained  in  a;  a 

or,  what  part  a  is  of  b.  b 

I.  If  b  be  contained  an  exact  number  of  times  in  a  their 
ratio  is  a  whole  number. 

If  b  be  not  contained  an  exact  number  of  times  in  a,  but 

if  there  be  a  common  measure  which  is  contained  m  times  in  a 

m 
and  n  times  in  b,  their  ratio  is  the  fraction  — . 

II.  If  a  and  b  be  incommensurable,  their  ratio  cannot  be 
exactly  expressed  in  figures.  But  if  b  be  divided  into  n  equal 
parts,  and  one  of  these  parts  be  contained  m  times  in  a  with 

a  remainder  less  than  -  part  of  b,  then  —  is  an  approximate 
n  n 

value  of  the  ratio  — ,  correct  within  - . 
6  n 


THEORY    OF    LIMITS.  87 


Again,  if  each  of  these  equal  parts  of  b  be  divided  into  n 
equal  parts ;  that  is,  if  b  be  divided  into  n2  equal  parts,  and  if 
one  of  these  parts  be  contained  m'  times  in  a  with  a  remainder 

1  m! 

less  than  —  part  of  b,  then  — ^  is  a  nearer  approximate  value 

of  the  ratio  - ,  correct  within  —^ . 
b  n 

By  continuing  this   process,   a   series  of  variable   values, 

—  >  — o  >  — ?  j  e^c->  w^u  be  obtained,  which  will  differ  less  and 
n       nl      ir 

less  from  the  exact  value  of  - .     We  may  thus  find  a  fraction 

which  shall  differ  from  this  exact  value  by  as  little  as  we  please, 
that  is,  by  less  than  any  assigned  quantity. 

Hence,  an  incommensurable  ratio  is  the  limit  toward  which 
its  successive  approximate  values  are  constantly  tending. 

On  the  Theory  of  Limits. 

196.  Dep.  When  a  quantity  is  regarded  as  having  a  fixed 
value,  it  is  called  a  Constant ;  but,  when  it  is  regarded,  under 
the  conditions  imposed  upon  it,  as  having  an  indefinite  number 
of  different  values,  it  is  called  a  Variable. 

197.  Def.  When  it  can  be  shown  that  the  value  of  a  vari- 
able, measured  at  a  series  of  definite  intervals,  can  by  indefinite 
continuation  of  the  series  be  made  to  differ  from  a  given  con- 
stant by  less  than  any  assigned  quantity,  however  small,  but 
cannot  be  made  absolutely  equal  to  the  constant,  that  constant 
is  called  the  Limit  of  the  variable,  and  the  variable  is  said  to 
approach  indefinitely  to  its  limit. 

If  the  variable  be  increasing,  its  limit  is  called  a  superior 
limit ;  if  decreasing,  an  inferior  limit. 

198.  Suppose  a  point  ± * u[      M"     B 

to  move  from  A  toward  B,  under  the  conditions  that  the  first  sec- 
ond it  shall  move  one-half  the  distance  from  A  to  B,  that  is, 
to  M;  the  next  second,  one-half  the  remaining  distance,  that  is, 
to  M' ;  the  next  second,  one-half  the  •remaining  distance,  that 
is,  to  M",  and  so  on  indefinitely. 

Then  it  is  evident  that  the  moving  point  may  approach  as 
near  to  B  as  we  please,  but  will  never  arrive  at  B.     For,  however 


88  GEOMETRY. — BOOK    II. 

near  it  may  be  to  B  at  any  instant,  the  next  second  it  will  pass 
over  one-half  the  interval  still  remaining ;  it  must,  therefore, 
approach  nearer  to  B,  since  half  the  interval  still  remaining  is 
some  distance,  but  will  not  reach  B,  since  half  the  interval  still 
remaining  is  not  the  whole  distance. 

Hence,  the  distance  from  A  to  the  moving  point  is  an  in- 
creasing variable,  which  indefinitely  approaches  the  constant 
A  B  as  its  limit;  and  the  distance  from  the  moving  point  to  B 
is  a  decreasing  variable,  which  indefinitely  approaches  the  con- 
stant zero  as  its  limit. 

If  the  length  of  A  B  be  two  inches,  and  the  variable  be 
denoted  by  x,  and  the  difference  between  the  variable  and  its 
limit,  by  v : 

after  one  second,         x  =  l,  v  =  l 

after  two  seconds,       #  =  1  -f  £,  v  =  l 

after  three  seconds,     a;  =  1  -J-  $  +  £,  v  =  i 

after  four  seconds,       #  =  l  +  £  +  £-f-&,    v  =  i 
and  so  on  indefinitely. 
Now  the  sum  of  the  series  1  +  \ r  -f-  \  +  &  etc.,  is  evidently 
less  than  2 ;  but  by  taking  a  great  number  -of  terms,  the  sum 
can  be  made  to  differ  from  2  by  as  little  as  we  please.     Hence 
2  is  the  limit  of  the  sum  of  the  series,  when  the  number  of  the 
terms  is  increased  indefinitely  ;  and  0  is  the  limit  of  the  vari- 
able difference  between  this  variable  sum  and  2. 
Urn.  wilt  be  used  as  an  abbreviation  for  limit. 
199.  [1]    The  difference  between  a  variable  and  its  limit  is 
a  variable  whose  limit  is  zero. 

[2]  If  two  or  more  variables,  v,  v\  vn,  etc.,  have  zero  for  a 
limit,  their  sum,  v  -f-  v'-\-v",  etc.,  will  have  zero  for  a  limit. 

[3]  If  the  limit  of  a  variable,  v,  be  zero,  the  limit  of  a  ±  v 
will  be  the  constant  a,  and  the  limit  of  a  X  v  will  be  zero. 

[4]  The  product  of  a  constant  and  a  variable  is  also  a  va- 
riable, and  the  limit  of  the  product  of  a  constant  and  a  variable 
is  the  product  of  the  constant  and  the  limit  of  the  variable. 

[5]  The  sum  or  product  of  two  variables,  both  of  which  are 
either  increasing  or  decreasing,  is  also  a  variable. 


THEORY    OF    LIMITS.  89 


Proposition  I. 
[6]  If  two  variables  be  always  equal,  their  limits  are  equal. 

Let  the  two  variables  A  M  and 
A  N  be  always  equal,  and  let  A  C 
and  A  B  be  their  respective  limits. 

We  are  to  prove  A  C  =  A  B. 

Suppose  A  C  >  A  B.  Then  we  may 
diminish  A  C  to  some  value  A  C  such 
t\mtAC'=AB.  p 

Since  A  M  approaches  indefinitely  to   Ck 
A  C,  we  may  suppose  that  it  has  reached 
a  value  A  P  greater  than  A  C. 

Let  A  Q  be  the  corresponding  value  of  A  N. 

Then  AP=AQ. 

Now  A  C  =  A  B. 

But  both  of  these  equations  cannot  be  true,  for  A  P  >  A  C, 
and  A  Q  <  A  B.     .'.AC  cannot  be  greater  than  A  B. 

Again,  suppose  AC  <  A  B.  Then  we  may  diminish  A  B  to 
some  value  A  B'  such  that  A  C  =  A  B'. 

Since  A  X  approaches  indefinitely  to  A  B  we  may  suppose 
that  it  lias  reached  a  value  A  Q  greater  than  A  B'. 

Let  A  P  be  the  corresponding  value  of  A  M. 

Then  AP=AQ. 

Now  A  C  =  A  B'. 

But  both  of  these  equations  cannot  be  true,  for  A  P  <  A  C, 
and  A  Q>  A  B'.     .'.AC  cannot  be  less  than  A  B. 

Since  A  C  cannot  be  greater  or  less  than  A  B,  it  must  be 
equal  to  A  B.  Q-  E-  D- 

[7]  Corollary  1.  If  two  variables  be  in  a  constant  ratio, 
their  limits  are  in  the  same  ratio.    For,  let  x  and  y  be  two  variables 

x 

having  the  constant  ratio  r,  then  -  =  r,  or,  x  =  r  yf  therefore 

lim.  ( x) 
Urn.  (x)  =  lim.  (r  y)  =  rX  lim.  (y),  therefore  ,,       /  [  =  r. 

lim.  (y) 

[8]  Cor.  2.    Since  an  incommensurable  ratio  is  the  limit  of 

a 


its  successive  approximate  values,  two  incommensurable  ratios  -r 

a' 
and  —  are  equal  if  they  always  have  the  same  approximate  values 

when  expressed  ivitliin  the  same  measure  of  precision. 


90  GEOMETRY. BOOK  II. 

Proposition  II. 

[9]  The  limit  of  the  algebraic  sum  of  tivo  or  more  variables 
is  the  algebraic  sum  of  their  limits. 

Let  x,  y,  z,  be  variables,  a,  b,  and  c,    a - H~ 

their  respective  limits,  and  v,  v',  and  v", 

the  variable  differences  between  x,  y,  z,    b + — 

and  a,  b,  c,  respectively. 

We  are  to  prove  Urn.  (x  +  y  +  z)  =  a  +  b  +  c.     c *-— 

Now,  x  =  a  —  v,  y  =  b  —  v/,  z  =  c  —  v". 
Then,  x  +  y  +  z  =  a  —  v  +  b  —  v'  +  c  —  v". 

.'.  lim.(x~\- y-\-z)=lim.(a — v+b  —  v'  +  c — v").  [6] 

But,    Urn.  (a  —  v+b  —  v'  +  c  —  v")  =  a+b  +  c.  [3] 

.*.  Urn.  (x  +  y  +  z)  =  a  +  b  +  c. 

Q.  E.  D. 

Proposition  III. 

[10]  The  limit  of  the  product  of  two  or  more  variables  is  the 
product  of  their  limits. 

Let  x,  y,  z,  be  variables,  a,  b,  c,  their  respective 
limits,  and  v,  v',  v",  the  variable  differences  between 
x,  y,  z,  and  a,  b,  c,  respectively. 

We  are  to  prove  Urn.  (x  y  z)  —  ab  c. 
Now,  x  =  a  —  v,  y  —  b  —  v',  z  =  c  —  v". 
Multiply  these  equations  together. 

Then,  xy  z  =  ab  c=f=  terms  which  contain  one  or  more  of 
the  factors  v,  v',  v",  and  hence  have  zero  for  a  limit.  [3] 

.*.  Urn.  (xyz)  =  Urn.  (abc^f  terms  whose  limits  are  zero).  [6] 
But  Urn.  (a  b  c  =f  terms  whose  limits  are  zero)  =  ab  c. 
.  * .  Urn.  (xyz)  =  a  b  c. 

Q.  E.  D. 

For  decreasing  variables  the  proofs  are  similar. 


Note.  —  In  the  application  of  the  principles  of  limits,  refer- 
ence to  this  section  (§  199)  will  always  include  the  fundamental 
truth  of  limits  contained  in  Proposition  I.  ;  and  it  will  be  left  as 
an  exercise  for  the  student  to  determine  in  each  case  what  other 
truths  of  this  section,  if  any,  are  included  in  the  reference. 


MEASUREMENT   OP   ANGLES.  91 


Proposition  XII.     Theorem. 

200.  In  the  same  circle,  or  equal  circles,  two  commen- 
surable arcs  have  the  same  ratio  as  the  angles  which  thej/ 
subtend  at  the  centre. 


P 

In  the  circle  A  PC  let  the  two  arcs  be  A  B  and  A  C, 
and  AO B  and  AOC  the  A  which  they  subtend. 

m  .  arc  AB        ZAOB 

We  are  to  prove      =   -_ • 

F  arc  A  C        Z  AOC 

Let  H  K  be  a  common  measure  of  A  B  and  A  C. 
Suppose  //  K  to  be  contained  in  A  B  three  times, 
and  in  A  C  five  times. 

arc  AB       3 

Then  77^  =  7  • 

arc  A  C       5 

At  the  several  points  of  division  on  A  B  and  A  0  draw  radii. 

These  radii  will  divide  Z.  AOC  into  five  equal  parts,  of 

which  ZAOB  will  contain  three,  §  180 

(in  the  same  O,  or  equal  ©,  equal  arcs  subtend  equal  A  at  tlie  centre). 

ZAOB       3 


But 


' Z AOC       5 

arc  AB       3 

arc  AC       5 

arc  A  B      ZAOB 

Q 

Ax.  1. 

arc  AC       Z  AOC 

E.  D. 

92 


GEOMETRY. 


BOOK    II. 


Proposition  XIII.     Theorem. 

201.  In  the  same  circle,  or  in  equal  circles,  incom- 
mensurable arcs  have  the  same  ratio  as  the  angles  which 
they  subtend  at  the  centre. 

pt  P 


In  the  two  equal  ©  ABB  and  A'B'P1  let  AB  and  A' B' 
be  two  incommensurable  arcs,  and  C,  C  the  A  which 
they  subtend  at  the  centre. 

We  are  to  prove       =  . 

7  arc  AB         Z  0 

Let  A  B  be  divided  into  any  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  A'  B'  as  often  as  it  will  be 
contained  in  A'B'. 

Since  AB  and  A' B'  are  incommensurable,  a  certain  num- 
ber of  these  parts  will  extend  from  A1  to  some  point,  as  D, 
leaving  a  remainder  D  B'  less  than  one  of  these  parts. 
Draw  CD. 
Since  A  B  and  A'B  are  commensurable, 

arc_^/>  =  ZA'C'D  c  20Q 

arcAB     ~~   ZACB' 

(two  commensurable  arcs  have  the  same  ratio  as  the  A  which  they  subtend  at 

the  centre). 

Now  suppose  the  number  of  parts  into  which  A  B  is  divided 
to  be  continually  increased  ;  then  the  length  of  each  part  will 
become  less  and  less,  and  the  point  J)  will  approach  nearer  and 
nearer  to  B',  that  is,  the  arc  A'  D  will  approach  the  arc  A'  B'  as 
its  limit,  and  the  Z  A'  C  D  the  Z  A'  OB'  as  its  limit. 


MEASUREMENT    OP    ANGLES.  93 


Then  the  limit  of  ™a  A' D  will  he  alc  A' B' , 
arc  A  B  arc  A  B 

and  tlie  limit  of  4A'C'  D  will  be  Z  A'C'%. 
ZACB  ZACB 

Moreover,  the  corresponding  values  of  the  two  variables, 
namely, 

arcA'D       ,  Z  A'  C  D 

and 1 

arc  4  if  ZACB 

are  equal,  however  near  these  variables  approach  their  limits. 

.  .  their  limits  and are  equal.        6  199 

sltcAB  ZACB  X  * 

Q.  E.  D. 


202.  Scholium.  An  angle  at  the  centre  is  said  to  be  meas- 
ured by  its  intercepted  arc.  This  expression  means  that  an  angle 
at  the  centre  is  such  part  of  the  angular  magnitude  about  that 
point  (four  right  angles)  as  its  intercepted  arc  is  of  the  whole 
circumference. 

A  circumference  is  divided  into  360  equal  arcs,  and  each 
arc  is  called  a  degree,  denoted  by  the  symbol  (°). 

The  angle  at  the  centre  which  one  of  these  equal  arcs  sub- 
tends is  also  called  a  degree. 

A  quadrant  (one-fourth  a  circumference)  contains  there- 
fore 90°  ;  and  a  right  angle,  subtended  by  a  quadrant,  con- 
tains 90°. 

Hence  an  angle  of  30°  is  £  of  a  right  angle,  an  angle  of  45° 
is  £  of  a  right  angle,  an  angle  of  135°  is  f  of  a  right  angle. 

Thus  we  get  a  definite  idea  of  an  angle  if  we  know  the 
number  of  degrees  it  contains. 

A  degree  is  subdivided  into  sixty  equal  parts  called  min- 
utes, denoted  by  the  symbol  ('). 

A  minute  is  subdivided  into  sixty  equal  parts  called  sec- 
onds, denoted  by  the  symbol  ("). 


94 


GEOMETRY. 


BOOK    II. 


Proposition  XIV.     Theorem. 

203.  An  inscribed  angle  is  measured  by  one-half  of  the 
arc  intercepted  between  its  sides. 


Case  I. 

In  the  circle  PAB  {Fig.  1),  let  the  centre  C  be  in  one 
of  the  sides  of  the  inscribed  angle  B. 

We  are  to  prove      Z  B  is  measured  by  J  arc  PA. 

Draw  CA. 

CA  =  GB, 

(being  radii  of  the  same  O). 

.\ZB  =  ZA,  §112 

(being  opposite  equal  sides). 

ZPCA=ZB+ZA.  §105 

(the  exterior  Z  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

Substitute  in  the  above  equality  Z  B  for  its  equal  Z  A. 

Then  we  have       ZPCA  =  2ZB. 

But  Z  P  C  A  is  measured  by  A  P,  §  202 

(the  A  at  the  centre  is  measured  by  the  intercepted  arc). 

.'.  2  Z  B  is  measured  by  A  P. 

. ' .  Z  B  is  measured  by  \  A  P. 


MEASUREMENT    OF    ANGLES.  95 

Case  II. 

In    the    circle   BAE   {Fig.    2),    let   the    centre    C    fall 
within  the  angle  E  B  A. 

We  are  to  prove      Z  E  B  A  is  measured  by  \  arc  E  A. 

Draw  the  diameter  B  C  P. 

Z  P B A  is  measured  by  \  arc  PA,  (Case  I.) 

Z  P  B  E  is  measured  by  \  arc  P  E,  (Case  I.) 

.'.  Z  PBA  +  Z  PB E  is  measured  by  \  (arc  P4  +  arc  P E). 

.*.  Z  E  B  A  is  measured  by  J  arc  .#.4. 

Case  III. 

In  the  circle  BFP  (Fig.  3),  let  the  centre  C  fall  with- 
out the  angle  A  B  F. 

We  are  to  prove      Z  A  B  F  is  measured  by  J  arc  A  F. 

Draw  the  diameter  B  C  P. 

Z  P  B  F  is  measured  by  \  arc  P  F,  (Case  I.) 

Z  PBA  is  measured  by  \  arc  PA,  (Case  I.) 

.-.  Z  PBF—  Z  PBA  is  measured  by  J  (arc  PP  —  arc  PA). 

.'.  Z  ,4  2?  jF  is  measured  by  A  arc  ^1  i^. 

Q.  E.  D. 

204.  Corollary  1.  An  angle  inscribed  in  a  semicircle  is 
a  right  angle,  for  it  is  measured  by  one-half  a  semi-circumfer- 
ence, or  by  90°. 

205.  Cor.  2.  An  angle  inscribed  in  a  segment  greater  than 
a  semicircle  is  an  acute  angle ;  for  it  is  measured  by  an  arc  less 
than  one-half  a  semi-circumference  ;  i.  e.  by  an  arc  less  than  90°. 

206.  Cor.  3.  An  angle  inscribed  in  a  segment  less  than  a 
semicircle  is  an  obtuse  angle,  for  it  is  measured  by  an  arc  greater 
than  one-half  a  semi-circumference ;  i.  e.  by  an  arc  greater 
than  90°. 

207.  Cor.  4.  All  angles  inscribed  in  the  same  segment  are 
equal,  for  they  are  measured  by  one-half  the  same  arc. 


96  GEOMETRY. BOOK   II. 


Proposition  XV.     Theorem. 

208.  An  angle  formed  by  two  chords,  and  whose  vertex 
lies  between  the  centre  and  the  circumference,  is  measured  by 
one- half  the  intercepted  arc  plus  one-half  the  arc  intercepted 


by  its  sides  produced. 


Let  the  Z  AOC  be  formed  by  the  chords  A  B  and  CD. 

We  are  to  prove 

Z  A  0  C  is  measured  by  J  arc  A  C  +  \  arc  B  D. 

Draw  A  D. 

Z  COA=Z  D  +  ZA,  §105 

(the  exterior  Z  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A  ). 

But  Z  D  is  measured  by  \  arc  A  C,  §  203 

(an  inscribed  Z  is  measured  by  ?  the  intercepted  arc)  ; 

and  Z  A  is  measured  by  \  arc  B  D,  §  203 

.*.  Z  C '  0  A  is  measured  by  J  arc  A  C  +  J  arc  B  D. 

Q.  E.  D. 


Ex.  Show  that  the  least  chord  that  can  be  drawn  through 
a  given  point  in  a  circle  is  perpendicular  to  the  diameter  drawn 
through  the  point. 


MEASUREMENT    OF    ANGLES. 


97 


Proposition  XVI.     Theorem. 

209.  An   angle  formed   by  a  tangent   and  a  chord  is 
measured  by  one-half  the  intercepted  arc. 


Let  HAM  be  the  angle  formed  by  the   tangent  OM 
and  chord  AH. 

We  are  to  prove 

Z  HA  M  is  measured  by  £  arc  A  EH. 

Draw  the  diameter  AC  F. 

Z  FAMisa.it.  Z,  §186 

(the  radius  drawn  to  a  tangent  at  the  point  of  contact  is  A.  to  it). 

Z  FA  M,  being  a  rt.  Z,  is  measured  by  J  the  semi-circum- 
ference A  EF. 


Z  FA  H  is  measured  by  \  arc  FH, 
(an  inscribed  Z  is  measured  by  £  the  intercepted  arc) 


§203 


.'.  Z  FAM  —  A  FA H  is  measured  by  J  (arc  A  EF—  arc  HF). 
.'.  Z  HA  M  is  measured  by  J  arc  A  EH. 


Q.  E.  D. 


98 


GEOMETRY.  —  BOOK  II. 


Proposition  XYII.     Theorem. 

210.  An  angle  formed  by  two  secants,  two  tangents,  or 
a  tangent  and  a  secant,  and  which  has  its  vertex  without  the 
circumference,  is  measured  by  one-half  the  concave  arc,  minus 
one-half  the  convex  arc. 


M  D 

Fig.  1.  Fig.  2.  Fig.  3. 

Case  I. 
Let  the  angle  0  {Fig.  1)  be  formed  by  the  two  secants 
OA  and  OB. 

We  are  to  prove 

Z  0  is  measured  by  J  arc  AB  —  J  arc  E C. 
Draw  CB. 

ZACB  =  ZO  +  ZB,  §  105 

(the  exterior  A  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A  ). 

By  transposing, 

ZO  =  ZACB-ZB, 

But  Z  A  CB  is  measured  by  J  arc  A  B,  §  203 

(an  inscribed  Z  is  measured  by  £  the  intercepted  arc). 

and  Z  B  is  measured  by  \  arc  C  E,    *  §  203 

,\  Z  0  is  measured  by  J  arc  A  B  —  J  arc  CE, 


MEASUREMENT    OF    ANGIiES.  99 

Case  II. 

Let  the  angle    0  {Fig.  2)   be  formed  by  the  two  tan- 
gents OA  and  OB. 

We  are  to  prove 

Z  0  is  measured  by  £  arc  A  MB  —  £  arc  A  SB. 
Draw  A  B. 
ZABC  =  ZO  +  ZOAB}  §105 

(the  exterior  Z  of  a  A  is  equal  to  tfie  sum  oftJie  two  opposite  interior  A  ). 

By  transposing, 

Z  0=ZABC-Z  OAB. 

But       Z  ABC  is  measured  by  \  arc  A  MB,  §  209 

(an  Z  formed  by  a  tangent  and  a  chord  is  measured  by  £  the  intercepted  arc)y 

and         Z  0  A  B  is  measured  by  £  arc  A  SB.  §  209 

.'.  Z  0  is  measured  by  £  arc  A  MB  —  £  arc  A  SB. 

Case  III. 

Let  the  angle  0  {Fig.  3)    be  formed  by  the    tangent 
OB  and  the  secant  0 A. 

We  are  to  prove 

Z  0  is  measured  by  J  arc  A  D  S  —  J  arc  C  E  S. 
Draw  CS. 

ZACS  =  ZO  +  Z  CSO,  §  105 

(the  exterior  Zofak  is  equal  to  the  sum  of  the  two  opposite  interior  A). 
By  transposing, 

Z  0  =  ZACS~Z  CSO. 

But         Z  A  CS  is  measured  by  J  arc  A  D  S,  §  203 

(being  an  inscribcdZ). 

and         Z  CSO  is  measured  by  £  arc  C US,  §  209 

{being  an  Z  formed  by  a  tangent  and  a  chord). 

.'.  Z  0  is  measured  by  \  arc  A  D  S  —  J  arc  C E S. 

Q.  E.  D. 


100 


GEOMETRY. 


BOOK   II. 


Supplementary  Propositions. 
Proposition  XVIII.     Theorem. 

211.    Two  parallel  lines    intercept    upon   the  circum- 
ference equal  arcs. 

A 


Let  the  two  parallel  lines  C  A  and  B  F  (Fig.  1),  inter- 
cept the  arcs  C  B  and  A  F. 


We  are  to  prove      arc  C  B  =  arc  A  F. 

Draw  A  B. 

£A=£B, 

(being  alt.  -int.  A  ). 

But  the  arc  C  B  is  double  the  measure  of  Z.  A. 
and  the  arc  A  F  is  double  the  measure  of  A  B. 


68 


.*.  arc  C  B  =  arc  A  F. 


Ax.  (> 

Q.  E.  D. 


212.  Scholium.  Since  two  parallel  lines  intercept  on  the 
circumference  equal  arcs,  the  two  parallel  tangents  M N  and 
0  P  (Fig.  2)  divide  the  circumference  in  two  semi-circumferences 
AC  B  and  AQ  B,  and  the  line  A  B  joining  the  points  of  contact 
of  the  two  tangents  is  a  diameter  of  the  circle. 


SUPPLEMENTARY    PROPOSITIONS.  101 


Proposition  XIX.     Theorem. 

213.  If  the  sum  of  two  arcs  be  less  than  a  circum- 
ference the  greater  arc  is  subtended  by  the  greater  chord ; 
and  conversely,  the  greater  chord  subtends  the  greater  arc. 

B 


P 
In  the  circle  A  CP  let  the  two  arcs  A  B  and  BC  to- 
gether be  less   than  the  circumference,   and  let 
AB  be  the  greater. 

We  are  to  prove      chord  A  B  >  chord  B  C. 

Draw  A  C. 

In  the  A  A  B  C 

Z  C,  measured  by  J  the  greater  arc  AB,         §  203 

is  greater  than  Z  A,  measured  by  \  the  less  are  B  C. 

.'.  the  side  A  B  >  the  side  B  C,  §  117 

(in  a  A  the  greater  Z.  has  the  greater  side  opposite  to  it). 

Conversely  :     If   the    chord  A  B  be   greater  than  the 

chord  B  C. 

We  are  to  prove      arc  A  B  >  arc  B  C. 

In  the  A  A  B  C,- 

AB>BC,  Hyp. 

.-.ZOA,  §118 

(in  a  A  the  greater  side  has  the  greater  Z  opposite  to  it). 

.'.urcAB,  double  the  measure  of  the  greater  Z  C,  is  greater 
than  the  arc  B  C,  double  the  measure  of  the  less  Z  A. 

Q.  E.  D. 


102  GEOMETRY. BOOK    II. 


Proposition  XX.     Theorem. 

214.  If  the  sum  of  two  arcs  be  greater  than  a  circum- 
ference, the  greater  arc  is  subtended  by  the  less  chord ;  and, 
conversely,  the  less  chord  subtends  the  greater  arc. 

B 


E 
In  the  circle  BCE  let  the   arcs  AECB  and   BAEC 
together  be  greater  than  the  circumference,  and 
let  arc  AECB  be  greater  than  arc  B AEG. 

We  are  to  prove      chord  AB  <  chord  B  C. 

From  the  given  arcs  take  the  common  arc  AEC ; 

we  have  left  two  arcs,  CB  and  A  B,  less  than  a  circumference, 

of  which  CB  is  the  greater. 

.'.  chord  C  B  >  chord  A  B,  §  213 

(when  the  sum  of  two  arcs  is  less  than  a  circumference,  the  greater  arc  is 
subtended  by  the  greater  chord). 

.*.  the  chord  A  B,  which  subtends  the  greater  arc  AECB, 
is  less  than  the  chord  B C,  which  subtends  the  less  arc  BAE C. 

Conversely  :  If  the  chord  A  B  be  less  than  chord  B  C. 

We  are  to  prove      arc  AEC  B  >  arc  BAEC. 

Arc  AB  +  b,tcAECB  =  the  circumference. 

Arc  BC  +  arc  B  A  E  C  =  the  circumference. 
.*.  arc  A  B  +  arc  A  EC  B  =  arc  B  C  +  arc  BA  EC. 

But  arc  A  B  <  arc  B  C,  §  213 

(being  subtended  by  the  less  chord). 

.'.  &tcAECB>slicBAEC. 

Q.  E.  D. 


CONSTRUCTIONS.  103 


On  Constructions. 

Proposition  XXI.     Problem. 

215.  To  find  a  point  in  a  plane,  having  given  its  dis- 
tances from  two  known  points. 


X 


Let  A  and  B  be  the  two  known  points;  n  the  dis- 
tance of  the  required  point  from  At  o  its  distance 
from  B. 

It  is  required  to  find  a  point  at  the  given  distances  from  A 
and  B. 

From  A  as  a  centre,  with  a  radius  equal  to  n,  describe  an  arc. 

From  B  as  a  centre,  with  a  radius  equal  to  o,  describe  an  arc 
intersecting  the  former  arc  at  C. 

G  is  the  required  point. 

Q.  E.  F. 


216.  Corollary  1.  By  continuing  these  arcs,  another  point 
below  the  points  A  and  B  will  be  found,  which  will  fulfil  the 
conditions. 

217.  Cor.  2.  When  the  sum  of  the  given  distances  is  equal 
to  the  distance  between  the  two  given  points,  then  the  two  arcs 
described  will  be  tangent  to  each  other,  and  the  point  of  tan- 
gency  will  be  the  point  required. 


104  GEOMETRY. BOOK   II. 

Let  the  distance  from  A  to  B  equal  n  +  o. 

From  A  as  a  centre,  with  a  \j 

radius  equal  to  n,  describe  an  arc ;  A-  (p  -B 

and  from  B  as  a  centre,  with  A 

a  radius  equal  to  o,   describe  an 
arc.  * 

These  arcs  will  touch  each  ~ 

other  at  0,  and  will  not  intersect. 

.*.  G  is  the  only  point  which  can  be  found. 

218.  Scholium  1.  The  problem  is  impossible  when  the 
distance  between  the  two  known  points  is  greater  than  the  sum 
of  the  distances  of  the  required  point  from  the  two  given  points. 

Let  the  distance  from  A  to  B  be  greater  than  n  +  o. 

Then  from  A  as  a  centre, 
with  a  radius   equal  to  ?i,  de-  A'  '& 

scribe  an  arc; 

and  from  B  as  a  centre,  with  a 

radius  equal  to  o,  describe  an  arc. 

These  arcs  will  neither  touch 

o 
nor  intersect  each  other ; 

hence  they  can  have  no  point  in  common. 

219.  Scho.  2.  The  problem  is  impossible  when  the  distance 
between  the  two  given  points  is  less  than  the  difference  of  the 
distances  of  the  required  point  from  the  two  given  points. 

Let  the  distance  from  A  to  B  be  less  than  n  —  o. 

From  A  as  a  centre,  with  a  radius             ^**~~"      "^ 
equal  to  n,  describe  a  circle ;  /  \ 

and  from  B  as  a  centre,  with  a     j  /  \    \ 

radius  equal  to  o,  describe  a  circle.  j  /  \    \ 

The  circle  described  from  B  as  a     \  \  I   / 

centre  will  fall  wholly  within  the  circle      V  \v  y   / 

described  from  A  as  a  centre :  o  \  / 

hence  they  can  have  no  point  in  n > •-* 

common. 


n 


CONSTRUCTIONS.  105 


Proposition  XXII.     Problem. 

220.   To  bisect  a  given  straight  line. 
C 


X 

E 

Let  AB  be  the  given  straight  line. 
It  is  required  to  bisect  the  line  A  B. 

From  A  and  B  as  centres,  with  equal  radii,  describe  arcs 
intersecting  at  C  and  E. 

Join  OE. 
Then  the  line  C  E  bisects  A  B. 
For,  C  and  E,  being  two  points  at  equal  distances  from  the 
extremities  A  and  By  determine  the  position  of  a  J_  to  the  mid- 
dle point  of  A  B.  §  60 

Q.  E.  F. 

Proposition  XXIII.     Problem. 

221.  At  a  given  point  in  a  straight   line,  to   erect   a 
perpendicular  to  that  line.  R 


xix 


A 


HO" 
Let  0  be  the  given  point  in  the  straight  line  AB. 
It  is  required  to  erect  a  J_  to  the  line  A  B  at  the  point  0. 

TakeOH=OB. 
From  B  and  H  as  centres,  with  equal  radii,  describe  two 
arcs  intersecting  at  R. 

Then  the  line  joining  R  0  is  the  _L  required. 
For,  0  and  R  are  two  points  at  equal  distances  from  B  and  H,  and 
.*.  determine  the  position  of  a  JL  to  the  line  H B  at  its 
middle  point  0.  §  60 

Q.  E.  F, 


106  GEOMETRY.  —  BOOK   II. 


Proposition  XXIV.     Problem. 

222.  From  a  point  without  a  straight  line,  to  let  fall  a 
perpendicular  upon  that  line. 

C 


X 


V 


/ 


-^— 


H  \ m ^'K 

Let  AB  be  a  given  straight  line,  and  G  a  given  point 
without  the  line. 

It  is  required  to  let  fall  a  A.  to  the  line  A  B  from  the  point  G. 

From  G  as  a  centre,  with  a  radius  sufficiently  great, 

describe  an  arc  cutting  A  B  at  the  points  H  and  K. 

From  H  and  K  as  centres,  with  equal  radii, 

describe  two  arcs  intersecting  at  0. 

Draw  G  0, 

and  produce  it  to  meet  A  B  at  m. 

G  mid-  the  _L  required. 

For,  G  and  0,  being  two  points  at  equal  distances  from  H 
and  K,  determine  the  position  of  a  ±  to  the  line  H  K  at  its 
middle  point.  §  60 

Q.  E.  F. 


CONSTRUCTIONS. 


107 


Proposition  XXV.     Problem. 

223.  To  construct  an  arc  equal  to  a  given  arc  whose 
centre  is  a  given  point. 

F — ^ 


X 


/f\ 


a< 


B  B> 

Let  C  be  the  centre  of  the  given  arc  A  B. 

It  is  required  to  construct  an  arc  equal  to  arc  A  B. 

Draw  CB,  CA,  and  A  B. 

From  C  as  a  centre,  with  a  radius  equal  to  CB, 

describe  an  indefinite  arc  B'  F. 

From  B'  as  a  centre,  with  a  radius  equal  to  chord  A  B, 

describe  an  arc  intersecting  the  indefinite  arc  at  A'. 

Then  arc  A'  B'  =  arc  A  B. 

draw  chord  A'  B'. 


For, 


and 


The  (D  are  equal, 
(being  described  with  equal  radii), 

chord  A'  B'  =  chord  A  B ; 

.'.  arc  A'  B'  =  arc  A  B, 
(in  equal  <D  equal  chords  subtend  equal  arcs). 


Cons. 
§  182 


Q.  E.  F. 


108  GEOMETRY.  —  BOOK  II. 


Proposition  XXVI.     Problem. 

224.  At  a  given  point  in  a  given  straight  line  to  con- 
struct an  angle  equal  to  a  given  angle. 

F 


\ 


a 


B  B' 

Let  C  be  the  given  point  in  the  given  line  C  B4,  and 
C  the  given  angle. 

It  is  required  to  construct  an  Z  at  C  equal  to  the  Z  C. 

Prom  C  as  a  centre,  with  any  radius  as  C  B, 

describe  the  arc  A  B,  terminating  in  the  sides  of  the  Z. 

Draw  chord  A  B. 

From  C  as  a  centre,  with  a  radius  equal  to  C  B, 

describe  the  indefinite  arc  B'  F. 

Prom  B'  as  a  centre,  with  a  radius  equal  to  A  B, 

describe  an  arc  intersecting  the  indefinite  arc  at  A'. 

Draw  A'  C. 
Then  Z.G'  =  Za 
For,  iomA'B'. 

The  (D  to  which  belong  arcs  A  B  and  A'  B'  are  equal, 
(being  described  with  equal  radii). 

and  chord  A'  B'  =  chord  A  B ;  Cons. 

.*.arc^/^/  =  arc^j5,  §  182 

(in  equal  ©  equal  chords  subtend  equal  arcs). 

.-.ZC'  =  ZC,  §180 

(in  equal  (D  equal  arcs  subtend  equal  A  at  the  centre). 

Q.  E.  F- 


CONSTRUCTIONS. 


109 


Proposition  XXVII.     Problem. 
225.  To  bisect  a  given  arc. 


7*V 


Let  A  OB  be  the  given  arc. 

It  is  required  to  bisect  the  arc  A  0  B. 

Draw  the  chord  A  B. 

From  A  and  B  as  centres,  with  equal  radii, 

describe  arcs  intersecting  at  E  and  C. 

Draw  EC. 

E  C  bisects  the  arc  A  OB. 

For,  E  and  C,  being  two  points  at  equal  distances  from 
A  and  B,  determine  the  position  of  the  J_  erected  at  the  middle 
of  chord  A  B ;  §  60 

and  a  _L  erected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.  §  184 

Q.  E.  F. 


110 


GEOMETRY. BOOK   II. 


Proposition  XXVIII.     Problem. 
226.   To  bisect  a  given  angle. 


<    C 


Let  A  EB  he  the  given  angle. 

It  is  required  to  bisect  Z  A  E  B. 

From  E  as  a  centre,  with  any  radius,  as  E  A, 

describe  the  arc  A  0  B,  terminating  in  the  sides  of  the  Z. 

Draw  the  chord  A  B. 

From  A  and  B  as  centres,  with  equal  radii, 

describe  two  arcs  intersecting  at  C. 

Join  EC. 

E  0  bisects  the  Z  E. 

For,  E  and  (7,  being  two  points  at  equal  distances  from  A  and 
B,  determine  the  position  of  the  -L  erected  at  the  middle  of 
AB.  §  60 

And  the  _L  erected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.  §  184 

.'.  arc  A  0  =  arc  0  B, 

.'.Z  AEC  =  Z  BEG,  §  180 

(in  the  same  circle  equal  arcs  subtend  equal  A  at  the  centre). 

Q.  E.  F. 


CONSTRUCTIONS.  Ill 


Proposition  XXIX.     Problem. 

227.   Through  a  given  point  to  draw   a   straight   line 
parallel  to  a  given  straight  line. 

E -^  H 


B 


Let  AB  be  the  given  line,  and  H  the  given  point. 

t 
It  is  required  to  draw  through  the  point  H  a  line  II  to  the 

line  A  B. 

Draw  HA,  making  the  Z  H AB. 

At  the  point  H  construct  ZAHE  =  ZHAB. 

Then  the  line  HE  is  II  to  A  B. 

For,  ZEHA=ZHAB;  Cons. 

.'.  HE  is  II  to  A  B,  §  69 

(whtn  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
line,  if  the  alt.  -int.  A  be  equal,  tlie  lines  are  parallel). 

Q.  E.  F. 


Ex.  1.  Find  the  locus  of  the  centre  of  a  circumference  which 
passes  through  two  given  points. 

2.  Find  the  locus  of  the  centre  of  the  circumference  of  a 
given  radius,  tangent  externally  or  internally  to  a  given  cir- 
cumference. 

3.  A  straight  line  is  drawn  through  a  given  point  A,  inter- 
secting a  given  circumference  at  B  and  C.  Find  the  locus  of 
the  middle  point  P  of  the  intercepted  chord  B  Q, 


II 52  GEOMETRY. BOOK   11. 

Proposition  XXX.     Problem. 

228.  Two  angles  of  a  triangle  being  given  to  find  the 

third. 

R 


8^ 

v  / 


E 


Let  A  and  B  be  two  given  angles  of  a  triangle. 

It  is  required  to  find  the  third  Z  of  the  A. 

Take  any  straight  line,  as  E  F,  and  at  any  point,  as  H. 

construct  Z  R  H  F  equal  to  Z  B, 

and  Z.  SHF  equal  to  Z  A. 

Then  Z  RH S  is  the  Z  required. 

For,  the  sum  of  the  three  A  of  a  A  =  2  rt.  A,         §  98 

and  the  sum  of  the  three  A  about  the  point  H,  on  the  same 
side  ofFF=2  rt.  A.  §  34 

Two   A   of   the   A    being    equal    to    two   A   about    the 
point   ff,  Cons. 

the  thhd  Z  of  the  A  must  be  equal  to  the  third  Z  about 
the  point  H. 

Q.  E.  F. 


CONSTRUCTIONS.  113 


Proposition  XXXI.     Problem. 

229.  Two  sides  and   the  included  angle  of  a  triangle 
being  given,  to  construct  the  triangle. 


K  ^r--- A  H 


E 


A 


Let  the  two   sides  of  the  triangle  be  E  and  Fy  and 
the  included  angle  A. 

It  is  required  to  construct  a  A  liaving  two  sides  equal  to  E 
and  F  respectively \  and  their  included  Z.  —  Z.  A. 

Take  HK  equal  to  the  side  F. 

At  the  point  H  draw  the  line  H  Mt 

making  the  Z  K H M  =  Z  A. 

On  H  M  take  HC  equal  to  E. 

Draw  C  K. 

Then  A  C  HK  is  the  A  required. 

Q.E.  F 


114  GEOMETRY.  —  BOOK   II. 


Proposition  XXXII.     Problem. 

230.  A  side  and  two  adjacent  angles  of  a  triangle  being 
given,  to  construct  the  triangle. 


0 

/*»n 

/      \ 

/ 

E 

/ 

/ 

\ 

.^>». 

Let  C  E  be  the  given  side,  A  and  B  the  given  angles. 

It  is  required  to  construct  a  A  having  a  side  equal  to  C  E, 
and  two  A  adjacent  to  that  side  equal  to  A  A  and  B  resjyectively. 

At  point  C  construct  an  A  equal  to  A  A. 

At  point  E  construct  an  A  equal  to  A  B. 

Produce  the  sides  until  they  meet  at  0. 

Then  A  C  0  E  is  the  A  required. 

Q.  E.  F. 

231.  Scholium.  The  problem  is  impossible  when  the  two 
given  angles  are  together  equal  to,  or  greater  than,  two  right 
angles. 


CONSTRUCTIONS.  115 


Proposition  XXXIII.     Problem. 

232.   The  three  sides  of  a  triangle  being  given,  to  con- 
struct the  triangle. 

\C  m 


AS-- 


B  o 

Let  the  three  sides  be  m,  n,  and  o. 

It  is  required  to  construct  a  A  having  three  sides  respectively, 
equal  to  m,  n,  and  o. 

Draw  A  B  equal  to  n. 

From  A  as  a  centre,  with  a  radius  equal  to  o, 

describe  an  arc ; 

and  from  B  as  a  centre,  with  a  radius  equal  to  m, 

describe  an  arc  intersecting  the  former  arc  at  C. 
Draw  CA  and  C  B. 
Then  A  C  A  B  is  the  A  required. 

Q.  E.  F. 

233.  Scholium.    The  problem  is  impossible  when  one  side 
is  equal  to  or  greater  than  the  sum  of  the  other  two. 


116 


GEOMETRY. BOOK   II. 


Proposition  XXXIV.     Problem. 

234.   The  hypotenuse  and  one  side  of  a  right  triangle 
being  given,  to  construct  the  triangle. 


X 

c 
/ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

1 

i 

B 
Let  m  be  the  given  side,  and  o  the  hypotenuse. 

It  is  required  to  construct  a  rt.  A  having  the  hypotenuse 
equal  o  and  one  side  equal  m. 

Take  A  B  equal  to  m. 

At  A  erect  a  _L,  A  X. 

From  B  as  a  centre,  with  a  radius  equal  to  o, 

describe  an  arc  cutting  A  X  at  G. 

Draw  CB. 

Then  A  C  A  B  is  the  A  required. 

Q.  E.  F 


CONSTRUCTIONS. 


117 


Proposition  XXXV.     Problem. 

235.   The  base,  the  altitude,  and  an  angle  at  the  base, 
of  a  triangle  being  given,  to  construct  the  triangle. 


Xr 


,^-R 


™« 


Let  o  equal  the  base,  m  the  altitude,  and  C  the  angle 
at  the  base. 

It  is  required  to  construct  a  A  having  the  base  equal  to  o> 
the  altitude  equal  to  m,  and  an  Z.  at  tlie  base  equal  to  0. 

Take  A  B  equal  to  o. 

At  the  point  A,  draw  the  indefinite  line  A  R, 

making  the  Z  BA  R  =  Z  C. 

At  the  point  A,  erect  a  A.  A  X  equal  to  m. 

From  X  draw  XS  II  to  A  By 

and  meeting  the  line  A  R  at  S. 

Draw  SB. 

Then  A  A  SB  is  the  A  required. 

Q.  E.  F. 


118  GEOMETRY. BOOK    II. 


Proposition  XXXVI.     Problem. 

236.   Two  sides  of  a  triangle  and  the  angle  opposite  one 
of  them  being  given,  to  construct  the  triangle. 

Case  I. 

When  the  given  angle  is  acute,  and  the  side  opposite  to  it  is  less  than 
the  other  given  side. 

D 


A 


// 


x       / 


\ 
\ 


\ y 


A^~-—i-—~-^r—^ 


Let  c  be  the  longer  and  a  the  shorter  given  side,  and 
A  A  the  given  angle. 

It  is  required  to  construct  a  A  having  two  sides  equal  to  a 
and  c  respectively,  and  the  Z.  opposite  a  equal  to  given  Z.  A. 
Construct  /.DAE  equal  to  the  given  A  A. 
On  AD  take  A  B  =  c. 
From  JB  as  a  centre,  with  a  radius  equal  to  a, 
describe  an  arc  intersecting  the  side  A  E  at  C  and  C". 

Draw  B  C  and  B  C". 
Then  both  the  A  A  B  C  and  A  B  C"  fulfil  the  conditions, 
and  hence  we  have  two  constructions. 

When  the  given  side  a  is  exactly  equal  to  the  _L  B  C,  there 
will  be  but  one  construction,  namely,  the  right  triangle  ABC. 

When  the  given  side  a  is  less  than  B  C,  the  arc  described 
from  B  will  not  intersect  A  E,  and  hence  the  problem  is  im- 
possible. 


CONSTRUCTIONS.  119 


Case  II. 
When  the  given  angle  is  acute,  right,  or  obtuse,  and  the  side  opposite 


Koiiis  greater  than  the  otJier  given  side. 

D 

\ 


A 


B 


£ 


/  \ 


/ 


■'   \\  n 


/   \\       /  i  \ 


\ 
\ 


\       \  /  \ 


\/ Y     Vg  ^4 1 A*: 


Fig.  1.  Fig.  2. 


\ 


s 

When  the  given  angle  is  obtuse. 
Construct  tho  Z  DAE  (Fig.  1)  equal  to  the  given  Z  S. 

Take  A  B  equal  to  a. 
From  B  as  a  centre,  with  a  radius  equal  to  c, 
describe  an  arc  cutting  E  A  at  C,  and  E  A  produced  at  C. 
Join  BCzjidB  C. 

Then  the  A  A  B  C  is  the  A  required,  and  there  is  only  one 
construction ;  for  the  A  A  BC  will  not  contain  the  given  Z  S. 

When  the  given  angle  is  acute,  as  angle  B  A  C!. 
There  is  only  one  construction,  namely,  the  A  B AC  (Fig.  1). 

When  the  given  Z.  is  a  right  angle. 

There  are  two  constructions,  the  equal  ABAC  and  BAC 
(Fig.  2).  Q.  E.  F. 

The  problem  is  impossible  when  the  given  angle  is  right  or 
obtuse,  if  the  given  side  opposite  the  angle  be  less  than  the 
other  given  side.  §  117 


120  GEOMETRY. BOOK   II. 


Proposition  XXXVII.     Problem. 

237.  Two  sides  and  an  included  angle  of  a  parallelo- 
gram being  given,  to  construct  the  parallelogram. 
R 


i 


/ 


Let  m  and  o  be    the    two  sides,  and  C  the  included 
angle. 

It  is  required  to  construct  a  O  having  two  adjacent  sides 
equal  to  m  and  o  respectively,  and  their  included  /.  equal  to  Z.  C. 
Draw  A  B  equal  to  o. 
From  A  draw  the  indefinite  line  A  R, 
making  the  Z.  A  equal  to  Z  C. 
On  A  R  take  A  H  equal  to  m. 
From  H  as  a  centre,  with  a  radius  equal  to  o,  describe 
an  arc. 

From  B  as  a  centre,  with  a  radius  equal  to  my 

describe  an  arc,  intersecting  the  former  arc  at  E. 
Draw  EH  and  E B. 
The  quadrilateral  A  B E His  the  O  required. 
For,  AB  =  HE,  Cons. 

AH  =  BEt  Cons. 

.*.  the  figure  A  B  E  H  is  a  O,  §  136 

(a  quadrilateral,  which  has  its  opposite  sides  equal,  is  a  O ). 

Q.  E.  F. 


CONSTRUCTIONS.  121 


Proposition  XXXVIII.     Problem. 
238.   To  describe  a  circumference   through  three  points 
not  in  the  same  straight  line. 


/ 
/ 
/ 

/ 
/ 

/ 

1 
1 

1 

0 

\ 
\ 
\ 
\ 
\ 
\ 
\ 
\ 
1 
\ 

.-■ 

•'   '•    \ 

"■*•..         ' 

\    .•'"'' 

•       ! 

V(7 

*\\ 

..--'  / 

Let  the  three  points  be  A,  B,  and  C. 

It  is  required  to  describe  a  circumference  through  tJie  three 
points  Ay  By  and  C. 

Draw  A  B  and  B  C. 

Bisect  A  B  and  B  C. 

At  the  points  of  bisection,  E  and  F,  erect  J§  intersect- 
ing at  0. 

From  0  as  a  centre,  with  a  radius  equal  to  0  A,  describe  a 
circle. 

Q  ABC  is  the  O  required. 

For,  the  point  0,  being  in  the  _L  E  0  erected  at  the  middle 
of  the  line  A  B,  is  at  equal  distances  from  A  and  B ; 

and  also,  being  in  the  J_  F  0  erected  at  the  middle  of  the 
line  C  B,  is  at  equal  distances  from  B  and  C,  §  58 

{every  point  in  the  _L  erected  at  the  middle  of  a  straight  line  is  at  equal 
distances  from  the  extremities  of  that  line). 

.'.  the  point  0  is  at  equal  distances  from  A,  B,  and  C, 
and  a  O  described  from  0  as  a  centre,  with  a  radius  equal 
to  0  A,  will  pass  through  the  points  A,  B,  and  C. 

Q.  E.  F. 

239.  Scholium.  The  same  construction  serves  to  describe 
a  circumference  which  shall  pass  through  the  three  vertices  of  a 
triangle,  that  is,  to  circumscribe  a  circle  about  a  given  triangle. 


122 


GEOMETRY. 


BOOK    II. 


Proposition  XXXIX.     Problem. 

240.   Through  a  given  point  to  draw  a   tangent  to  a 
given  circle. 


>\E 


Case   1 .  —  When  tlie  given  point  is  on  the  circumference. 

Let  ABC  (Fig.  1)  be  a  given  circle,  and  G  the  given 
point  on  the  circumference. 

It  is  required  to  draw  a  tangent  to  the  circle  at  C. 

From  the  centre  0,  draw  the  radius  0  C. 

At  the  extremity  of  the  radius,  C,  draw  C  M  J_  to  0  C. 

Then  C  M  is  the  tangent  required,  §  186 

(a  straight  line  A.  to  a  radius  at  its  extremity  is  ta.ngent  to  the  O). 

Case   2.  —  When  the  given  point  is  without  the  circumference. 

Let  ABC  (Fig.  2)   be    the  given  circle,  0  its  centre, 
E  the  given  point  without  the  circumference. 
It  is  required  to  draw  a  tangent  to  the  circle  ABC  from 

the  point  E. 

Join  0  E. 

On  0  E  as  a  diameter,  describe  a  circumference  intersecting 
the  given  circumference  at  the  points  M  and  H. 

Draw  0  M  and  0 II,  EM  and  EH. 

Now  Z  OMEisa.xt.  Z,  §204 

(being  inscribed  in  a  semicircle). 

.'.  E M  is  _L  to  0  M  at  the  point  M; 

.'.EM  is  tangent  to  the  O,  §  186 

(a  straight  line  ±  to  a  radius  at  its  extremity  is  tangent  to  the  O). 
In  like  manner  we  may  prove  HE  tangent  to  the  given  O. 

Q.  E.  F. 

241.  Corollary.  Two  tangents  drawn  from  the  same  point 
to  a  circle  are  equal. 


CONSTRUCTIONS.  123 


Proposition     XL.     Problem. 

242.   To  inscribe  a  circle  in  a  given  triangle. 

fi 
M 


II 

Let  ABG  be  the  given  triangle. 

It  is  required  to  inscribe  a  O  in  the  A  A  B  G. 

Draw  the  line  A  E,  bisecting  Z  A, 

and  draw  the  line  G  E,  bisecting  Z.  G. 

Draw  EH  A.  to  the  line  A  C. 

From  E,  with  radius  EH,  describe  the  O  K M H. 

The  O  KHM  is  the  O  required. 

For,  draw  EK  ±to  A  B, 

&ndEM±toBG. 

In  the  rt.  A  A  KE  and  A  HE 

AE  =  AE,  Iden. 

ZEAK  =  Z.EAH,  Cons. 

.\AAKE  =  A  AHE,  §  110 

(Two  rt.  A  are  equal  if  the  hypotenuse  and  an  acute  Z  of  the  one  be  equal 
respectively  to  the  hypotenuse  and  an  acute  Z  of  the  other). 

.'.  EK=EH, 
(being  homologous  sides  of  equal  A). 

In  like  manner  it  may  be  shown  E M=  EH. 
.'.EKy  EH,  and  E M are  all  equal. 
.*.  a  O  described  from  E  as  a  centre,  with  a  radius  equal  to  EH, 

will  touch  the  sides  of  the  A  at  points  H,  K,  and  M,  and 
be  inscribed  in  the  A.  §  174 

Q.  E.  F. 


124 


GEOMETRY. 


BOOK    II. 


Proposition  XLI.     Problem. 


243.    Upon  a  given  straight  line,  to  describe  a  segment 
which  shall  contain  a  given  angle. 


7 


M 


Let  AB  be  the  given  line,  and  M  the  given  angle. 

It  is  required  to  describe  a  segment  upon  the  line  A  B,  which 
shall  contain  Z  M. 

At  the  point  B  construct  Z  ABE  equal  to  Z  M. 
Bisect  the  line  A  B  by  the  ±  F  H. 
Prom  the  point  B,  draw  B  0  _L  to  EB.    ■ 
Prom  0,  the  point  of  intersection  of  F H  and  B  0,  as  a 
centre,  with  a  radius  equal  to  0  B,  describe  a  circumference. 

Now  the  point  0,  being  in  al  erected  at  the  middle  of 
A  B,  is  at  equal  distances  from  A  and  B,  §  58 

(every  point  in  a  J_  erected  at  the  middle  of  a  straight  line  is  at  equal  dis- 
tances from  the  extremities  of  that  line)  ; 

.\  th«  circumference  will  pass  through  A. 

Now  B E is  ±  to  OB,  Cons. 

.'.BE  is  tangent  to  the  O,  §  186 

(a  straight  line  A.  to  a  radius  at  its  extremity  is  tangent  to  the  G). 

.'.  Z  A  B  E  is  measured  by  \  arc  A  B,  §  209 

(being  an  Z  formed  by  a  tangent  and  a  chord). 

Also  any  Z  inscribed  in  the  segment  A  II B,  as  for  instance 
Z  A  KB,  is  measured  by  J  arc  A  B,  §  203 

(being  an  inscribed  Z  ). 


CONSTRUCTIONS. 


125 


.\Z  AKB  =  ZABE, 

(being  both  measured  by  \  the  same  arc)  ; 
.'./.AKB  =  Z  M. 
segment  A  H  B  is  the  segment  required. 


Q.  E.  F. 


Proposition  XLII.     Problem. 

244.    To  find  the  ratio  of  two  commensurable  straight 

lines. 

E  H 

A » LJL-B 

K 


v. 1 r 

F 
Let  AB  and  C  D  be  two  straight  lines. 

It  is  required  to  find  the  greatest  common  measure  of  A  B 
and  C  D,  so  as  to  express  tlieir  ratio  in  figures. 

Apply  C  D  to  A  B  as  many  times  as  possible. 

Suppose  twice  with  a  remainder  E  B. 

Then  apply  E  B  to  C  D  as  many  times  as  possible. 

Suppose  three  times  with  a  remainder  F D. 
Then  apply  F D  to  E  B  as  many  times  as  possible. 

Suppose  once  with  a  remainder  H  B. 
Then  apply  H  B  to  F  D  as  many  times  as  possible. 

Suppose  once  with  a  remainder  K  D. 
Then  apply  K  D  to  H B  as  many  times  as  possible. 
Suppose  K  D  is  contained  just  twice  in  H  B. 
The  measure  of  each  line,  referred  to  K  D  as  a  unit,  will 
then  be  as  follows  :  — 

HB  =2KD; 

FD  =  IIB+  KD  =  3KD 
EB  =  FD  +  HB  =  5  KD 
CD  =3EB+  FD  =  18KD 
AB=20D+EB  =  41  KD. 

.    AJB  =     41  KD  , 

"  CD         ISKD' 

/.the  ratio  of  -—=  —  . 

6  °       18  Q.  E.  F. 


126  geometry. book  h. 

Exercises. 

1.  If  the  sides  of  a  pentagon,  no  two  sides  of  which  are 
parallel,  be  produced  till  they  meet ;  show  that  the  sum  of  all 
the  angles  at  their  points  of  intersection  will  be  equal  to  two 
right  angles. 

2.  Show  that  two  chords  which  are  equally  distant  from  the 
centre  of  a  circle  are  equal  to  each  other ;  and  of  two  chords,  that 
which  is  nearer  the  centre  is  greater  than  the  one  more  remote. 

3.  If  through  the  vertices  of  an  isosceles  triangle  which  has 
each  of  the  angles  at  the  base  double  of  the  third  angle,  and  is 
inscribed  in  a  circle,  straight  lines  be  drawn  touching  the  circle ; 
show  that  an  isosceles  triangle  will  be  formed  which  has  each 
of  the  angles  at  the  base  one-third  of  the  angle  at  the  vertex. 

4.  A  D  B  is  a  semicircle  of  which  the  centre  is  0 ;  and  AEG 
is  another  semicircle  on  the  diameter  AC',  A  T  is  a  common 
tangent  to  the  two  semicircles  at  the  point  A.  Show  that  if 
from  any  point  F,  in  the  circumference  of  the  first,  a  straight 
line  FG  be  drawn  to  G,  the  part  FK,  cut  off  by  the  second 
semicircle,  is  equal  to  the  perpendicular  FH  to  the  tangent  A  T. 

5.  Show  that  the  bisectors  of  the  angles  contained  by  the 
opposite  sides  (produced)  of  an  inscribed  quadrilateral  intersect 
at  right  angles. 

6.  If  a  triangle  A  B  0  be  formed  by  the  intersection  of  three 
tangents  to  a  circumference  whose  centre  is  0,  two  of  which, 
A  M  and  A  N,  are  fixed,  while  the  third,  B  G,  touches  the  cir- 
cumference at  a  variable  point  P ;  show  that  the  perimeter  of 
the  triangle  A  B  G  is  constant,  and  equal  to  A  M  +  A  N,  or 
2  AM.     Also  show  that  the  angle  B  0  G  is  constant. 

7.  A  B  is  any  chord  and  A  G  is  tangent  to  a  circle  at  A, 
G  D  E  a  line  cutting  the  circumference  in  D  and  E  and  parallel 
to  A  B ;  show  that  the  triangle  A  G  D  is  equiangular  to  the 
triangle  E  A  B. 


CONSTRUCTIONS.  127 


Constructions. 

1.  Draw  two  concentric  circles,  such  that  the  chords  of  the 
outer  circle  which  touch  the  inner  may  be  equal  to  the  diameter 
of  the  inner  circle. 

2.  Given  the  base  of  a  triangle,  the  vertical  angle,  and  the 
length  of  the  line  drawn  from  the  vertex  to  the  middle  point 
of  the  base  :  construct  the  triangle. 

3.  Given  a  side  of  a  triangle,  its  vertical  angle,  and  the  radius 
of  the  circumscribing  circle  :  construct  the  triangle. 

4.  Given  the  base,  vertical  angle,  and  the  perpendicular  from 
the  extremity  of  the  base  to  the  opposite  side  :  construct  the 
triangle. 

5.  Describe  a  circle  cutting  the  sides  of  a  given  square,  so 
that  its  circumference  may  be  divided  at  the  points  of  inter- 
section into  eight  equal  arcs. 

6.  Construct  an  angle  of  60°,  one  of  30°,  one  of  120°,  one 
of  150°,  one  of  45°,  and  one  of  135°. 

7.  In  a  given  triangle  ABC,  draw  Q  D  E  parallel  to  the  base 
B  C  and  meeting  the  sides  of  the  triangle  at  D  and  E,  so  that 
D E  shall  be  equal  to  DB  +  EC. 

8.  Given  two  perpendiculars,  A  B  and  CD,  intersecting  in  0, 
and  a  straight  line  intersecting  these  perpendiculars  in  E  and  F ; 
to  construct  a  square,  one  of  whose  angles  shall  coincide  with 
one  of  the  right  angles  at  O,  and  the  vertex  of  the  opposite  angle 
of  the  square  shall  lie  in  E  F.     (Two  solutions.) 

9.  In  a  given  rhombus  to  inscribe  a  square. 

10.  If  the  base  and  vertical  angle  of  a  triangle  be  given ; 
find  the  locus  of  the  vertex. 

11.  If  a  ladder,  whose  foot  rests  on  a  horizontal  plane  and 
top  against  a  vertical  wall,  slip  down ;  find  the  locus  of  its 
middle  point. 


BOOK  III. 

PROPORTIONAL  LINES  AND  SIMILAR  POLYGONS. 


Ox  the  Theory  of  Proportion. 

245.  Def.  The  Terms  of  a  ratio  are  the  quantities  com- 
pared. 

246.  Def.    The  Antecedent  of  a  ratio  is  its  first  term. 

247.  Def.    The  Consequent  of  a  ratio  is  its  second  term. 

248.  Def.  A  Proportion  is  an  expression  of  equality  be- 
tween two  equal  ratios. 

A  proportion  may  he  expressed  in  any  one  of  the  following 
forms :  — 

1.  a  :  b  :  :  c  :  d 

2.  a  :  b  =  c  :  d 

3.  a-=c-. 
b       d 

Form  1  is  read,  a  is  to  b  as  c  is  to  d. 

Form  2  is  read,  the  ratio  of  a  to  6  equals  .the  ratio  of  c  to  d. 

Form  3  is  read,  a  divided  by  b  equals  c  divided  by  d. 

The  Terms  of  a  proportion  are  the  four  quantities  com- 
pared. 

The  first  and  third  terms  in  a  proportion  are  the  ante- 
cedents, the  second  and  fourth  terms  are  the  consequents. 

249.  The  Extremes  in  a  proportion  are  the  first  and  fourth 
terms. 

250.  The  Means  in  a  proportion  are  the  second  and  third 
terms. 


THEORY    OF    PROPORTION.  129 

251.  Def.  Ill  the  proportion  a  :  b  :  :  c  :  d ;  d  is  a  Fourth 
Proportional  to  a,  b,  and  c. 

252.  Def.  In  the  proportion  a  :  b  :  :  b  :  c ;  c  is  a  Third 
Proportional  to  a  and  b. 

253.  Def.  In  the  proportion  a  :  b  :  :  b  :  c;  b  is  a  .Afea/i 
Proportional  between  a  and  c. 

254.  Def.  Four  quantities  are  Reciprocally  Proportional 
when  the  first  is  to  the  second  as  the  reciprocal  of  the  third  is  to 
the  reciprocal  of  the  fourth. 

Thus  a  :  b  :  :  -  :  - . 

c     d 

If  we  have  two  quantities  a  and  b,  and  the  reciprocals  of 

these  quantities  -  and  -  ;  these  four  quantities  form  a  recipro- 
a  b 

cal  proportion,  the  first  being  to  the  second  as  the  reciprocal  of 
the  second  is  to  the  reciprocal  of  the  first. 

As  a  :  b  :  :  I  :  - . 

b      a 

255.  Def.  A  proportion  is  taken  by  Alternation,  when  the 
means,  or  the  extremes,  are  made  to  exchange  places. 

Thus  in  the  proportion 

a  :  b  :  :  o  :  d, 
we  have  either 

a  :  c  :  :  b  :  d,     or,     d  :  b  :  :  c  :  a. 

256.  Def.  A  proportion  is  taken  by  Inversion,  when  the 
means  and  extremes  are  made  to  exchange  places. 

Thus  in  the  proportion 

a  :  b  :  :  c  :  d, 
by  inversion  we  have 

b  :  a  :  :  d  :  c. 

257.  Def.  A  proportion  is  taken  by  Composition,  when 
the  sum  of  the  first  and  second  is  to  the  second  as  the  sum  of 


130  GEOMETRY. BOOK   III. 

the  third  and  fourth  is  to  the  fourth ;  or  when  the  sum  of  the 
first  and  second  is  to  the  first  as  the  sum  of  the  third  and  fourth 
is  to  the  third. 


Thus  if                     a  :  b  : 

:  c  :  d, 

we  have  by  composition, 

a  +  b  :  b  : 

:  c  +  d 

:d, 

or,                      a  +  b  :  a  : 

:  c  +  d 

:  c. 

258.  Def.  A  proportion  is  taken  by  Division,  when  the 
difference  of  the  first  and  second  is  to  the  second  as  the  dif- 
ference of  the  third  and  fourth  is  to  the  fourth ;  or  when  the 
difference  of  the  first  and  second  is  to  the  first  as  the  difference 
of  the  third  and  fourth  is  to  the  third. 

Thus  if  a  :  b  :  :  c  :  d, 

we  have  by  division 

a  —  b  :  b  ::  c  —  d  :  dt 

or,  a  —  b  :  a  :  :  c  —  d  :  c, 

Proposition  I. 

259.    In  every  proportion  the  product  of  the  extremes  is 
equal  to  the  product  of  the  weans. 

Let  a  :  b  :  :  c  :  d. 

We  are  to  prove       ad  =  be. 


Now 

a 
b  = 

c 
d' 

whence, 

by  multiplyi 

ngby 

bd, 

ad  = 

--be. 

Q.  E.  D 


THEORY    OF    PROPORTION.  131 

In  the  treatment  of  proportion,  it  is  assumed  that  fractions 
may  be  found  which  will  represent  the  ratios.  It  is  evident  that 
a  ratio  may  be  represented  by  a  fraction  when  the  two  quanti- 
ties compared  can  be  expressed  in  integers  in  terms  of  any 
common  unit.  Thus  the  ratio  of  a  line  2£  inches  long  to  a  line 
3 \  inches  long  may  be  represented  by  the  fraction  §§  when  both 
lines  are  expressed  in  terms  of  a  unit  TV  of  an  inch  long. 

But  it  often  happens  that  no  unit  exists  in  terms  of  which 
both  the  quantities  can  be  expressed  in  integers.  In  such  cases, 
however,  it  is  possible  to  find  a  fraction  that  will  represent  the 
ratio  to  any  required  degree  of  accuracy. 

Thus,  if  a  and  b  denote  two  incommensurable  lines,  and  b  be 
divided  into  any  integral  number  (n)  of  equal  parts,  if  one  of 
these  parts  be  contained  in  a  more  than  m  times,  but  less  than 

m  + 1   times,   then     -  >  —  but  <  — — —  ;    so    that   the   error 
b        n  n 

in   taking   either   of    these   values    for  -is  <  —     Since  n  can 

b  n 

be  increased  at  pleasure,  -  can  be  made  less  than  any  assigned 
n 

value  whatever.    Propositions,  therefore,  that  are  true  for  —  and 

n 

i — — — ,  however  little  these  fractions  differ  from  each  other,  are 
n 

true  for  - ;  and  -  may  be  taken  to  represent  the  value  of  — 
b  n  b 


Proposition  II. 

260.  A   mean  proportional   between    two   quantities   is 
equal  to  the  square  root  of  their  product. 

In  the  proportion  a  :  b  :  :  b  :  c, 

62  =  a  c,  §  259 

(the  product  of  the  extremes  is  equal  to  the  product  of  the  means). 

Whence,  extracting  the  square  root, 
b  =  \fac 

Q.  E.  D. 


132  GEOMETRY. BOOK    III. 

Proposition  III. 

261.  If  the  product  of  two  quantities  be  equal  to  the 
product  of  two  others,  either  two  may  be  made  the  extremes 
of  a  proportion  in  which  the  other  two  are  made  the  means. 

Let  ad  =  be. 

We  are  to  prove      a  :  b  :  :  c  :  d. 

Divide  both  members  of  the  given  equation  by  b  d. 


a         c 
b~~d' 


Then 

or,  a  :  b  :  :  c  :  d. 


Q.  E.  D. 


Proposition  IV. 

262.    If  four  quantities  of  the  same  hind  be  in  propor- 
tion, they  will  be  in  proportion  by  alternation. 

Let  a  :  b  :  :  c  :  d. 
We  are  to  prove      a  :  c  :  :  b  :  d. 


kt  a        c 

Now,  -  =  -. 

6       d 

Multiply  each  member  of  the  equation  by  - 

c 


Then  ?  =  * 

c         a 


or,  a  :  c  :  :  b  :  d. 

Q.  E.  D. 


THEORY    OF    PROPORTION.  133 


Proposition  V. 

263.  If  four  quantities  be  in  proportion,  they  will  be  in 
proportion  by  inversion. 

Let  a  :  b  :  :  c  :  d. 

We  are  to  prove     b  :  a  :  :  d  :  c. 

Now,  £w5. 

b    d 

Divide  1  by  each  member  of  the  equation. 

Then  !  =  *, 

a        c 

or,  b  :  a  :  :  d  :  c. 

Q.  E.  o. 


Proposition  VI. 

264.  If  four  quantities  be  in  proportion,  they  will  be  in 
proportion  by  composition. 

Let  a  :  b  :  :  c  :  d 


We  are 

to  prove 

a  +  b 

:  b  :  :  c  +  d  : 

<L 

Now 

a 
b 

c 

Add  1  to  each  member  of  the  equation. 

Then 

that  is. 

a 

a  +  b 

c 

=  5+1' 

c  +  d 

b  d 

a  +  b  :  b  :  :  c  +  d  :  d. 


Q.  E    D 


134  GEOMETRY. BOOK    III. 

Proposition  VII. 

265.  If  font  quantities  be  in  proportion,  they  will  be  in 
proportion  by  division. 

Let  a  :  b  :  :  c  :  d. 
We  are  to  prove      a  —  b  :  b  :  :  c  —  d  :  d. 

Now  *=1. 

b        d 
Subtract  1  from  each  member  of  the  equation. 

Then 
that  is, 


or, 


a               c 

a  —  b        c  —  d 

b               d    ' 

a  —  b  :  b  :  :  c  —  d  : 

d. 

Q.  E.  D. 


Proposition  VIII. 
266.    In  a  series  of  equal  ratios,  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  any  antecedent  is 
to  its  consequent. 

Let  a  :  b  =  c  :  d  —  e  :f  =  g  :  h. 
We  are  to  prove      a  +  c  +  e  +  g  :  b  +  d  +  f  +  h  :  :  a  :  b. 
Denote  each  ratio  by  r. 

a         c        e        g 
Then  »— j-3  =/  =  T 

Whence,     a  =  br,       c  =  dr,       e  —fr,       g  =  hr. 
Add  these  equations. 

Then        a  +  c  +  e  +  g  =  (b  +  d  +  f  +  h)  r. 
Divide  by  (b  +  d  +  /  +  h). 

Then  a  +  c+g  +  -*gara-a> 

b+d+f+k      r       b 

or,     a  +  c  +  e  +  g  :  b  +  d  +  /  +  h  :  :  a  :  b. 

Q.  E.  D. 


THEORY    OF    PROPORTION.  135 

Proposition  IX. 
267.    The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 


Let  a  :  b  :  :  c  : 

* 

e  if  :  :  g  : 

K 

k  :  I  :  :  m  : 

n} 

We  art 

•  to  'prove      aek  :  bfl  i  : 

cgm 

:  dhn. 

Now 

a         c          e        g 

k  = 
J 

m 
n 

Whence  by  multiplication, 

aek        cgm 
bfl         dhn 

t 

or, 

aek  :  bfl  :  :  cgm 

:  dh 

n. 

Q. 

E.  D. 

Proposition  X. 

268. 

Like  powers,  or  like  roots 

j  of  the  terms 

• 

a  j&ra- 

portion  are 

in  proportion. 

Let  a  :  b  :  :  c 

:  d. 

We  are 

to  prove      an  :  bn  :  :  cn 

:  d», 

and 

a*  :  6n  :  :  C*  : 

dn. 

Now 

a        c 
b  =  d' 

By  raising  to  the  n01  power, 

—  =        :  or  an   :  bn 
bn        dn 

:  :  c* 

:  d» 

By  extracting  the  wth  root, 

l            i 
a»         c»              1,1 
=  —  :  or,  an  :  6* 

i            i 

i 
:  :c» 

i 
.  d*. 

6*»        c?n 

Q.  E.  D. 

269.  Def.  Equimultiples  of  two  quantities  are  the  products 
obtained  by  multiplying  each  of  them  by  the  same  number. 
Thus  m  a  and  m  b  are  equimultiples  of  a  and  b. 


136  GEOMETRY. BOOK    III. 


Proposition  XI. 

£70.  Equimultiples   of  two  quantities  are  in  the  same 

ratio  as  the  quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 

We  are  to  prove      ma  :  mb  :  :  a  :  b. 

at  a         a 

Now  _  =  _  . 

b         b 
Multiply  both  terms  of  first  fraction  by  m. 

mi  m  a         a 

Then  = 

mb         b 

or,  ma  :  mb  :  :  a  :  b. 

Q.  E.  D. 

Proposition  XII. 

271.  If  two  quantities  be  increased  or  diminished  by 
like  parts  of  each,  the  results  will  be  in  the  same  ratio  as  the 
quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 

We  are  to  prove      a  ±  —  a  :  b  ±  I    b  :  :  a  :  b. 
q  q 

In  the  proportion, 

ma  :  mb  :  :  a  :  b} 

substitute  for  m,  1   ±  -  . 
9. 

Then         (l  ±  i\  a  :  (l  ±  E\  b  :  :  a  :  b, 


a  ±P.a  :  b  ±?b 


Q.  E.  D. 


272.  Dep.    Euclid's  test  of  a  proportion  is  as  follows  :  — 
"  The  first  of  four  magnitudes  is  said  to  have  the  same  ratio 
to  the  second  which  the  third  has  to  the  fourth,  when  any  equi- 
multiples whatsoever  of  the  first  and  third  being  taken,  and  any 
equimultiples  whatsoever  of  the  second  and  fourth ; 


THEORY    OF    PROPORTION.  137 

"  If  the  multiple  of  the  first  be  less  than  that  of  the  second, 
the  multiple  of  the  third  is  also  less  than  that  of  the  fourth ;  or, 

"  If  the  multiple  of  the  first  be  equal  to  that  of  the  second, 
the  multiple  of  the  third  is  also  equal  to  that  of  the  fourth ;  or, 

"  If  the  multiple  of  the  first  be  greater  than  that  of  the 
second,  the  multiple  of  the  third  is  also  greater  than  that  of 
the  fourth." 


Proposition  XIII. 

273.  If  four  quantities  be  proportional  according  to  the 
algebraical  definition,  they  will  also  be  proportional  according 
to  Euclid's  definition. 

Let  a,  b,  c,  d  be  proportional  according  to  the  alge- 

a        c 
braical  definition ;  that  is  -r=  -j- 

We  are  to  prove  a,  6,  c,  d,  proportional  according  to  Euclid '$ 
definition. 

Multiply  each  member  of  the  equality  by  —  . 

n 

r™  ma        mc 

Then  —  = 

nb         n  d 

Now  from  the  nature  of  fractions, 

if  m a  be  less  than  nb,  mc  will  also  be  less  than  n d ; 

if  m a  be  equal  to  nb,  mc  will  also  be  equal  to  n d ; 

if  m a  be  greater  than  nb,  mc  will  also  be  greater  than  n d. 

.'.  a,  b,  c,  d  are  proportionals  according  to  Euclid's  def- 
inition. 

Q.  E.  D. 


138  GEOMETRY. BOOK   III. 


Exercises. 

1.  Show, that  the  straight  line  which  bisects  the  external 
vertical  angle  of  an  isosceles  triangle  is  parallel  to  the  base. 

2.  A  straight  line  is  drawn  terminated  by  two  parallel 
straight  lines ;  through  its  middle  point  any  straight  line  is 
drawn  and  terminated  by  the  parallel  straight  lines.  Show  that 
the  second  straight  line  is  bisected  at  the  middle  point  of  the 
first. 

3.  Show  that  the  angle  between  the  bisector  of  the  angle  A 
of  the  triangle  ABC  and  the  perpendicular  let  fall  from  A  on 
BG  is  equal  to  one-half  the  difference  between  the  angles  B 
and  C. 

4.  In  any  right  triangle  show  that  the  straight  line  drawn 
from  the  vertex  of  the  right  angle  to  the  middle  of  the  hypote- 
nuse is  equal  to  one-half  the  hypotenuse. 

5.  Two  tangents  are  drawn  to  a  circle  at  opposite  extremities 
of  a  diameter,  and  cut  off  from  a  third  tangent  a  portion  A  B. 
If  C  be  the  centre  of  the  circle,  show  that  A  C  B  is  a  right  angle. 

6.  Show  that  the  sum  of  the  three  perpendiculars  from  any 
point  within  an  equilateral  triangle  to  the  sides  is  equal  to  the 
altitude  of  the  triangle. 

7.  Show  that  the  least  chord  which  can  be  drawn  through  a 
given  point  within  a  circle  is  perpendicular  to  the  diameter 
drawn  through  the  point. 

8.  Show  that  the  angle  contained  by  two  tangents  at  the 
extremities  of  a  chord  is  twice  the  angle  contained  by  the  chord 
and  the  diameter  drawn  from  either  extremity  of  the  chord. 

9.  If  a  circle  can  be  inscribed  in  a  quadrilateral ;  show  that 
the  sum  of  two  opposite  sides  of  the  quadrilateral  is  equal  to  the 
sum  of  the  other  two  sides. 

10.  If  the  sum  of  two  opposite  sides  of  a  quadrilateral  be 
equal  to  the  sum  of  the  other  two  sides;  show  that  a  circle 
can  be  inscribed  in  the  quadrilateral, 


PROPORTIONAL    LINES.  139 


On  Proportional  Lines. 

Proposition  I.     Theorem. 

274.  If  a  series  of  parallels  intersecting  any  two 
straight  lines  intercept  equal  parts  on  one  of  these  lines, 
they  will  intercept  equal  parts  on  the  other  also. 

H  W 


K  K> 

Let  the  series  of  parallels  A  A',  B  B',  CC,  D  D',  E  E', 
intercept  on  H' K'  equal  parts  A'B',  B'C,  CD',  etc. 

We  are  to  prove 

they  intercept  on  H  K  equal  parts  A  B,  B  C,  C  D,  etc. 

At  points  A  and  B  draw  A  m  and  B  n  II  to  H'  K'. 

Am  =  A'B',  §135 

(parallels  comprehended  between  parallels  are  equal). 

Bn  =  B'C,  §  135 

.'.  A  m  =  Bn. 
In  the  A  B  Am  and  C  B  v, 

ZA=ZB,  §  77 

(having  their  sides  respectively  II  and  lying  in  the  same  direction  from 
the  vertices). 

Z  m  =  Z  n,  §  77 

and  Am  =  Bn, 

.'.  ABAm  =  A  CBn,  §  107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other). 

.*.  AB  =  BC, 

(being  homologous  sides  of  equal  A). 
In  like  manner  we  may  prove  BC  =  CD,  etc. 

Q.  E.  D. 


140 


GEOMETRY.  —  BOOK   III. 


Proposition  II.     Theorem. 

275.  If  a  line  be  drawn  through  two  sides  of  a  triangle 
parallel  to  the  third  side,  it  divides  those  sides  propor- 
tionally. 


Fig.  1.  Fig.  2. 

In  the  triangle  ABC  let  E F  be  drawn  parallel  to  B C. 

xrr  v  EB        FC 

We  are  to  prove        =  . 

AE        AF 

Case  I.  —  When  A  E  and  EB  (Fig.  1)  are  commensurable. 

Find  a  common  measure  of  A  E  and  E  B,  namely  B  m. 

Suppose  B  m  to  be  contained  in  B  E  three  times, 

and  in  A  E  five  times. 

Then  ££«?: 

AE        5 

At  the  several   points  of  division  on  B  E  and  A  E  draw 
straight  lines  II  to  B  C. 

These  lines  will  divide  A  C  into  eight  equal  parts, 

of  which  FC  will  contain  three,  and  A  F  will  contain  five,    §  274 

(if  parallels  intersecting  any  two  straight  lines  intercept  equal  parts  on  one 

of  these  lines,  they  will  intercept  equal  parts  on  the  other  also). 

.    FC  __  3 

"  AF~  5' 

EB       3 

AE       5' 

.    EB  =  FC 

'  '  AE       TF' 


But 


Ax.  1 


PROPORTIONAL    LINES.  141 


Case.   II.  —  When  A  E  and  E  B  (Fig.  2)  are  incommensurable. 

Divide  A  E  into  any  number  of  equal  parts, 

and  apply  one  of  these  parts  to  EB  as  often  as  it  will  be 
contained  in  E  B. 

Since  A  E  and  E  B  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  E  to  a  point  K,  leaving  a  re- 
mainder KB,  less  than  one  of  the  parts. 

Draw  KB  II  to  BC. 
Since  A  E  and  E  K  are  commensurable, 

EK       FH  in       T. 

AE  =  AF  (CaS6L) 

Suppose  the  number  of  parts  into  which  A  E  is  divided  to 
be  continually  increased,  the  length  of  each  part  will  become  less 
and  less,  and  the  point  K  will  approach  nearer  and  nearer  to  B. 

The  limit  of  E K  will  be  E B,  and  the  limit  of  FH  will  be  FC 

.*.  the  limit  of will  be  , 

AE  AE 

and  the  limit  of  will  be 

AF  AF 

•pi  jr  ~p  it 

Now  the  variables  and  — —  are  always  equal,   how- 

A  E  AF 

ever  near  they  approach  their  limits  ; 

A  their  limits  EJL  and  L£-  are  equal,  §  199 

A&  A  r 

Q.  E.  D. 

276.  Corollary.  One  side  of  a  triangle  is  to  either  part 
cut  off  by  a  straight  line  parallel  to  the  base,  as  the  other  side  is 
to  the  corresponding  part. 

Now  EB  :  AE  :  :  FC  :  AF.  §  275 

By  composition, 

EB  +  A  E  :  A  E  :  :  FC  +  A  F  :  A  F,  §  263 

or,  A  B  :  A  E  :  :  A  C  :  A  F. 


142  GEOMETRY. BOOK   III. 


Proposition  III.     Theorem. 

277.  If  a  straight  line  divide  two  sides  of  a   triangle 
proportionally,  it  is  parallel  to  the  third  side. 

A 


In  the  triangle  ABC  letEF  be  drawn  so  that  —  =  —  . 

AF     AF 

We  are  to  prove      F  F  II  to  B  C. 

From  F  draw  E  H  \\  to  B  C. 

(one  side  of  a  A  is  to  either  part  cut  off  by  a  line  II  to  the  base,  as  the  other 
side  is  to  the  corresponding  part). 

But  44  -  M-  HyP- 


Ax.  1 


.*.  F F  and  F II  coincide, 
(their  extremities  being  the  same  points). 

But  FH  is  II  to  BC;  Cons. 

.*.  F F,  which  coincides  with  F II,  is  II  to  BC. 

Q.  E.  D. 

278.  Def.  Similar  Polygons  are  polygons  which  have  their 
homologous  angles  equal  and  their  homologous  sides  proportional. 

Homologous  points,  lines,  and  angles,  in  similar  polygons, 
are  points,  lines,  and  angles  similarly  situated. 


AF 

AF 

.    AC 
'  '  AF 

AC 
AH' 

.'.  AF  = 

AH. 

SIMILAR    POLYGONS.  143 


On  Similar  Polygons. 
Proposition  IV.     Theorem. 
279.    Two  triangles  which  are  mutually  equiangular  are 

similar. 

A 
A'  A 


In  the  A  ABC  and  A1  B1  C  let  A  A,  B,  C  be  equal  to 

A  A',  B',  C  respectively. 

We  are  to  prove      A  B  :  A'  B'  =  AC  :  A'  C  =  BC  :  B'  C. 
Apply  the  A  A' B1  C  to  the  A  ABC, 
so  that  Z  A'  shall  coincide  with  Z  A. 

Then  the  A  A'  B'  C  will  take  the  position  of  A  A  E  H. 

Now  Z  A  EH  (same  as  Z  B')  =  Z  B. 

.'.  EH  is  II  to  BC,  §  69 

{when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
line,  if  the  ext.  int.  A  be  equal  the  lines  are  parallel), 

.'.AB:AE  =  AC:AH,  §276 

(one  side  of  a  A  is  to  either  part  cut  off  by  a  line  II  to  tlie  base,  as  tlie  other 
side  is  to  the  corresponding  part). 

Substitute  for  A  E  and  A  H  their  equals  A'  B'  and  A'  C. 

Then  AB  :  A< B>  =  AC  :  A'C. 

In  like  manner  we  may  prove 

A  B  :  A'  B'  =  B  C  :  B'  C. 

.*.  the  two  A  are  similar.  §  278 

Q.  E.  D. 

280.  Cor.   1.    Two  triangles  are  similar  when  two  angles 
of  the  one  are  equal  respectively  to  two  angles  of  the  other. 

281.  Cor.  2.    Two  right  triangles  are  similar  when  an  acute 
angle  of  the  one  is  equal  to  an  acute  angle  of  the  other. 


144 


GEOMETRY. 


BOOK    III. 


Proposition  V.     Theorem. 
282.    Two    triangles   which   have  their   sides  respectively 
proportional  are  similar. 


In  the  triangles  ABG  and  A' B' C  let 
AB         AG         BG 
J^B'  ~~  A'  G'  ~~  B'O' 
We  are  to  prove 
A  A,  B,  and  G  equal  respectively  to  A  A',  B'y  and  C'. 

Take  on  A  B,  A  E  equal  to  A'  B1, 
and  on  AG,  AH  equal  to  A'  &,     Draw  EH. 
AB         AG 
A'B>        A'Gr 
Substitute  in  this  equality,  for  A'  B'  and  A'  G'  their  equals 


Hyp. 


A  E  and  A  H. 
Then 


AB        AG 
AE       AH' 

.\EH  is  II  to.SC,  • 
{if  a  line  divide  two  sides  of  a  A  proportionally,  it  is 

Now  in  the  A  A  BG  and  A  EH 

Z  ABG  =  ZAEH, 

(being  ext.  int.  angles). 

Z  ACB  =  Z  A  HE, 

Z  A=  Z  A. 

.-.  A  AB  G  and  A  EH  are  similar, 
(two  mutually  equiangular  A  are  similar). 

;   AB       AB 

' "  BG  ~  EH1 

(homologous  sides  nf  simi'ar  A  are  proportional). 


§277 

to  the  third  side). 

§  70 

§70 

Iden. 
§  279 

§278 


SIMILAR    POLYGONS. 


145 


But 


AB 
BC 
AE 
EH 


A'B' 


Hyp- 
Ax.  1 


Cons. 


§108 


A'B' 
B'G1' 
Since  A  E  —  A'  B', 

EH  =  B'C. 
Now  in  the  AAEH  and  A'  B'  C, 

EH=B'C,  AE  =  A'B',  and  A  H  =  A' Cf, 
.-.A  AEH=AA'B'C, 
(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

But  A  A  E  H  is  similar  to  A  ABC. 

.-.  A  A'  B'  C  is  similar  to  A  ABC. 

Q.  E.  D. 

283.  Scholium.  The  primary  idea  of  similarity  is  likeness 
of  form ;  and  the  two  conditions  necessary  to  similarity  are  : 

I.  For  every  angle  in  one  of  the  figures  there  must  be  an 
equal  angle  in  the  other,  and 

II.  the  homologous  sides  must  be  in  proportion. 

In  the  case  of  triangles  either  condition  involves  the  other, 
but  in  the  case  of  other  polygons,  it  does  not  follow  that  if  one 
condition  exist  the  other  does  also. 


W 


R 


Thus  in  the  quadrilaterals  Q  and  Q',  the  homologous  sides 
are  proportional,  but  the  homologous  angles  are  not  equal  and 
the  figures  are  not  similar. 

In  the  quadrilaterals  R  and  R',  the  homologous  angles  are 
equal,  but  the  sides  are  not  proportional,  and  the  figures  are  not 
similar. 


146  GEOMETRY. BOOK   III. 

Proposition  VI.     Theorem. 

284.   Two  triangles  having  an  angle  of  the  one  equal  to 

an  angle  of  the  other,  and  the  including  sides  proportional, 

are  similar. 

A 
Af  /\ 


In  the  triangles  ABC  and  A' B' C  let 
£A=  Z.A',   and 


A'B1        A'C 

We  are  to  prove      A  A  B  C  and  A'  B'  C  similar. 

Apply  the  A  A' B' C  to  the  A  ABC  so  that  Z  A'  shall 
coincide  with  Z.  A. 

Then  the  point  B'  will  fall  somewhere  upon  A  B,  as  at  E, 

the  point  C  will  fall  somewhere  upon  A  0,  as  at  H,  and 

B' C  upon  E  H. 

at  AB         AC  jj 

Now  = Hyp. 

A'B'        A'C  JF 

Substitute  for  A'  B'  and  A1  C  their  equals  A  E  and  A  H. 

Then  i*^^. 

AE       AH 

.'.the  line   EH  divides  the  sides   AB  and  AC  propor- 
tionally ; 

.'.EH  is  II  to  BC,  §  277 

(if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  !l  to  the  third  side). 

.'.  the  A  A  BC  and  A  E  H  are  mutually  equiangular  and  similar. 

,'.  A  A'B' C  is  similar  to  A  ABC. 

Q.  E.  D. 


SIMILAR   POLYGONS.  147 


Proposition  VII.     Theorem. 

285.    Two  triangles  which  have  their  sides  respectively 
parallel  are  similar. 


In  the  triangles  ABC  and  A' B'  C  let  AB.AC,    and 
BC  be  'parallel  respectively   to  A' B'y  A'C,  and 
B'C. 
We  are  to  prove      A  A  B  C  and  A'  B'  C  similar. 

The  corresponding  A  are  either  equal,  §  77 

(two  A  ichose  sides  arc  II,  two  and  two,  and  lie  in  the  same  direction,  or 
opposite  directions,  from  their  vertices  are  equal). 

or  supplements  of  each  other,  §  78 

(if  two  A  have  two  sides  II  and  lying  in  the  same  direction  from  their  vertices, 

while  the  other  two  sides  are  II  and  lie  in  opposite  directions,  the  A  are 

supplements  of  each  other). 

Hence  we  may  make  three  suppositions : 

1st.  A  +  A'  =  2rt.A,    B  +  B'  =  2vt.A,     (7+C"  =  2rt.  A. 
2d.  A  =  Af,  B  +  B'  =  2vt.A,     C  +  C  =  2 rt.  A. 

3d.  A=A',  B  =  B>  .'.  C=C. 

Since  the  sum  of  the  A  of  the  two  A  cannot  exceed  four 
right  angles,  the  3d  supposition  only  is  admissible.  §  98 

.'.  the  two  A  A  B  C  and  A'  B'  C  are  similar,         §  279 

(two  mutually  equiangular  A  are  similar). 

Q.  E.  D. 


148 


GEOMETRY. BOOK    III. 


Proposition  VIII.     Theorem. 

286.   Two  triangles  which  have  their  sides  respectivt 
perpendicular  to  each  other  are  similar. 

B 


In  the  triangles  EFD  and  B  A  C,  let  E F,  FD  and  ED, 
be  perpendicular  respectively  to  AC,'BG  and  A  B. 

We  are  to  prove  A  E ''F D  and  B  AC  similar. 

Place  the  A  E  FD  so  that  its  vertex  E  will  fall  on  A  B, 
and  the  side  E  F,  JL  to  A  C,  will  cut  A  C  at  F'. 

Draw  F'  D'  II  to  F  D,  and  prolong  it  to  meet  B  C  at  H. 
In  the  quadrilateral  B  E  D'H,  JL  E  and  //  are  rt.  A . 

.-.ZB  +  ZED'  H=2  rt.  A. 


But  ZED'  F'  +  Z  ED  11=2  rt.  A. 

.'.ZED'  F'  =  ZB. 

Now  ZC+ZHF'C=ri.Z, 

(in  a  rt.  A  the  sum  of  the  two  acute  A  =  a  rt.  Z) 
and  ZEF'D'  +  Z  IIF'C  =  rt.  Z. 

.'.ZEF'D'  =  ZC. 
.'.AEF'D'  and  B  AC  are  similar. 
But  A  EFD  is  similar  to  A  E F  D'. 

.'.  A  E  F  D  and  B  A  C  are  similar. 


§158 

§34 

Ax.  3. 

§103 

Ax.  9. 
Ax.  3. 

§280 
§279 

Q.  E.  D. 


287.  Scholium.  When  two  triangles  have  their  sides  re- 
spectively parallel  or  perpendicular,  the  parallel,  sides,  or  the 
perpendicular  sides,  are  homologous. 


SIMILAR   POLYGONS."  149 


Proposition  IX.     Theorem. 

288.  Lines  drawn  through  the  vertex  of  a  triangle  divide 
proportionally  the  base  and  its  parallel. 


In  the  triangle  ABC  let  II L  be  parallel  to  AC,  and 
let  BS  and  BT  be  lines  drawn  through  its  ver- 
tex to  the  base. 


We 

are 

to  prove 

AS 
HO 

= 

ST 

on 

= 

TC 
RL 

A  B  HO  and  B  AS  are  similar,  §  279 
(two  &  which  are  mutually  equiangular  are  similar). 

A  B  0  R  and  B  S  T  are  similar,  §  279 

A  B  R  L  and  B  T  C  are  similar,  §  279 

•    AJL        (^\        $?  _  (BT\        TC  §278 

"  HO        \0 B/  ~~~  0 R  ~\B RJ  ~~     RL'  S  - 


(homologous  sides  of  similar  &  are  proportional). 


Q.  E.  D. 


Ex.  Show  that,  if  three  or  more  non-parallel  straight  lines 
divide  two  parallels  proportionally,  they  pass  through  a  common 
point. 


150  GEOMETRY. BOOK   III. 


Proposition  X.     Theorem. 

289.  If  in  a  right  triangle  a  perpendicular  be  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse  : 

I.  It  divides  the  triangle  into  two  right  triangles  which 
are  similar  to  the  whole  triangle,  and  also  to  each  other: 

II.  The  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse. 

III.  Each  side  of  the  right  triangle  is  a  mean  pro- 
portional between  the  hypotenuse  and  its  adjacent  segment. 

IV.  The  squares  on  the  two  sides  of  the  right  triangle 
have  the  same  ratio  as  the  adjacent  segments  of  the  hypote- 
mise. 

V.  The  square  on  the  hypotenuse  has  the  same  ratio  to 
the  square  on  either  side  as  the  hypotenuse  has  to  the  segment 
adjacent  to  that  side. 

B 


-        F 

In  the  right  triangle  ABC,  let  B F  be  drawn  from  the 
vertex  of  the  right  angle  B,  perpendicular  to  the 
hypotenuse  A  C. 

I.    We  are  to  prove 

the  AABF,  ABC,  and  FBC  similar. 
In  the  rt.  A  BA  F  and  BA  C,  . 

the  acute  Z.  A  is  common. 

.*.  the  A  are  similar,  §  281 

{two  rt.  A  are  similar  when  an  acute  Z  of  the  one  is  equal  to  an  acute  Z 
of  the  other). 

In  the  rt.  ABCFqm&BCA, 

the  acute  Z  C  is  common. 

.*.  the  A  are  similar.  §  281 

Now  as  the  rt.  AABF  and  C  B  F  are  both  similar  to 
A  B  C,  by  reason  of  the  equality  of  their  A, 

they  are  similar  to  each  other. 


SIMILAR   POLYGONS.  151 


II.  We  are  to  prove      A  F  :  BF  :  :  BF  :  FO. 
In  the  similar  A  A  B  F  and  G  B  F, 

A  F,  the  shortest  side  of  the  one, 
B  F,  the  shortest  side  cf  the  other, 
B  F,  the  medium  side  of  the  one, 
F  G,  the  medium  side  of  the  other. 

III.  We  are  to  prove      A  G  :  A  B  :  :  A  B  :  A  F. 
In  the  similar  A  A  B  G  and  A  B  F, 

A  G,  the  longest  side  of  the  one, 
A  B,  the  longest  side  of  the  other, 
A  B,  the  shortest  side  of  the  one, 
A  F,  the  shortest  side  of  the  other. 
Also  in  the  similar  A  A  B  C  aiid  F  B  G, 

A  C,  the  longest  side  of  the  one, 
B  C,  the  longest  side  of  the  other, 
B  C,  the  medium  side  of  the  one, 
F  G,  the  medium  side  of  the  other. 

™-     nr         ,  AT?        AF 

IV.  We  are  to  prove     =  . 

EC1       F<J 
In  the  proportion  A  G  :  A  B  :  :  A  B  :  A  J?, 

A~B2  =  AG  X  AF,  §  259 

(the  product  of  the  extremes  is  equal  to  the  product  of  the  means). 

and  in  the  proportion  AG  :  BG  :  :  BG  :  FG, 

F7?=-ACX  FG.  §259 

Dividing  the  one  by  the  other, 

JTB2        AGX  AF 

KJ?  ~  AGX  FG' 
Cancel  the  common  factor  A  G,  and  we  have 

re  _  af 

BG2        FG' 

V.  We  are  to  prove    HO.  =  ££. 

AW        ^F 
jfG2=ACX  AG. 

X22  =  AGX  AF,  (Case  III.) 

Divide  one  equation  by  the  other  : 

then  £?*  =    AGX  AG  _  AC 

J3f       AGXAF       AF  q.  e.  o. 


152  GEOMETRY. BOOK    III. 


Proposition  XI.     Theorem. 

290.  If  two  chords  intersect  each  other  in  a  circle,  their 
segments  are  reciprocally  proportional. 


Let   the    two    chords  AB   and  EF  intersect  at   the 
point  0. 

We  are  to  prove     AO  :  EO  :  :  OF  :  OB. 

Draw  ,4^  and  E  B. 

In  the  AAOFtmdEOB, 

ZF=ZB,  §203 

(each  being  measured  by  \  arc  A  E). 

A  A=*  A  E,  §  203 

(each  being  measured  by  £  arc  F  B). 

.*.  the  A  are  similar.  §  280 

(two  A  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  the  other). 

Whence      A  0,  the  medium  side  of  the  one,  §  278 

E  0,  the  medium  side  of  the  other, 
0  F,  the  shortest  side  of  the  one, 
0  B,  the  shortest  side  of  the  other. 

Q.  E.  D. 


SIMILAR    POLYGONS.  153 


Proposition  XII.     Theorem. 


291.  If  from  a  point  without  a  circle  two  secants  be 
drawn,  the  whole  secants  and  the  parts  without  the  circle 
are  reciprocally  proportional. 


Let  OB  and  OC  be  two  secants  drawn  from  point  0. 
We  are  to  prove      OB  :  OC  :  :  OM  :  OH. 

Draw  H  C  and  M  B. 
In  the  A  OIIC  and  OMB 

/.  0  is  common, 

Z  B  =  Z.  C,  §  203 

{each  being  measured  by  $  arc  H M). 

.'.  the  two  A  are  similar,  §  280 

(two  &  are  similar  wJicn  two  A  of  the  one  are  equal  to  two  A  of  the  other). 

Whence  0  B,  the  longest  side  of  the  one,  §278 

:  0  C,  the  longest  side  of  the  other, 

:  :  0  M,  the  shortest  side  of  the  one, 

:  0  If,  the  shortest  side  of  the  other. 

Q.  E.  D. 


154  GEOMETRY. BOOK   m. 


Proposition  XIII.     Theorem. 

292.  If  from  a  point  without  a  circle  a  secant  and  a 
tangent  he  drawn,  the  tangent  is  a  mean  proportional  between 
the  whole  secant  and  the  part  without  the  circle. 

0 


c 

Let  OB  be  a  tangent  and  0  C  a  secant  drawn  from 
the  point  0  to  the  circle  MBC. 

We  are  to  prove      0  C  :  0  B  :  :  0  B  :  0  M. 

Draw  BM  and  £  C. 

In  the  A  tf£i/and  OBG 

Z.  0  is  common. 

A  OBMis  measured  by  J  arc  MB,  §  209 

(being  an  Z  formed  by  a  tangent  and  a  chord). 

A  G  is  measured  by  \  arc  B  M,  §  203 

(being  an  inscribed  Z. ). 

.\Z  OBM=Z  C. 

.'.AOBCsmdOBMsive  similar,  §  280 

(having  tvjo  A  of  the  one  equal  to  two  A  of  the  other). 

Whence      0  C,  the  longest  side  of  the  one,  §  278 

0  B,  the  longest  side  of  the  other, 

0  B,  the  shortest  side  of  the  one, 

0  My  the  shortest  side  of  the  other. 

Q.  E.  D. 


SIMILAR   POLYGONS. 


155 


Proposition  XIV.     Theorem. 

293.  If  two  polygons  be  composed  of  the  same  number 
of  triangles  which  are  similar,  each  to  each,  and  similarly 
placed,  then  the  polygons  are  similar. 
E 


C  B'  C 

In  the  two  polygons  ABODE  and  A'B'O'D'E',  let 
the  triangles  B  A  E,  BEO,  and  CED  be  similar 
respectively  to  the  triangles  B'  A'  E'f  B'  E'  0\  and 
C'E'  D'. 

We  are  to  prove 
the  polygon  ABODE  similar  to  the  polygon  A'  B'  0'  D'  E'. 

Z  A=Z  A',  §  278 

(being  lunnologous  A  of  similar  A). 

Z  ABE  =  Z  A'B'E',  §278 

Z  EBO  =  Z  E'B'O',  §278 

Add  the  last  two  equalities. 

Then  Z  ABE+  Z  E B 0  =?  Z  A1  B' E>  +  Z  E' B' C ; 
or,  ZABO^ZA'B'C. 

In  like  manner  we  may  prove  Z  BO D  =  Z  B'  C  D'y  etc. 

.*.  the  two  polygons  are  mutually  equiangular. 
AE       A 


Now 


IB      /EB\_  BO  _(EO\_OD  ^ED 
A'E'     ArB,~~\ErB')~BlOl~\E'Cl)      CD1     E' D1' 
(the  homologous  sides  of  similar  A  are  proportional).       • 
.'.  the  homologous  sides  of  the  two  polygons  are  proportional. 

.*.  the  two  polygons  are  similar,  §  278 

(having  their  homologous  A  equal,  and  tlieir  homologous  sides  proportional). 

Q.  E.  D. 


156 


GEOMETRY. 


BOOK   III. 


Proposition  XV.     Theorem. 


294.  If  two  polygons  be  similar,  they  are  co 
(he  same  number  of  triangles,  which  are  similar  and 
pla 


of 
ilarly 


B  C  B'  a 

Let  the  polygons  ABODE  and  A'B'C D'E'  be  similar. 

From  two  homologous  vertices,  as  E  and  E', 

draw  diagonals  EB,  EC,  and  E< B',  E  C. 

We  are  to  prove      A  A  E  B,  EBG,  EC  D 

similar  respectively  to  A  A'  E  B',  E'  B'  C,  E  C  D'. 

In  the  AAEB  and  A1  E' B1, 

Z  A=Z  A1,  §  278 

(being  homologous  A  of  similar  polygons). 

AE^=   AB_  §  278 

A'E'       A'B1' 

(being  homologous  sides  of  similar  polygons). 

.'.  A  A  E B  and  A'  E> B1  are  similar,  §  284 

(having  an  A  of  the  one  equal  to  an  A  of  the  oilier,  and  tJie  including 
sides  proportional). 

Also,  Z  ABC=Z  A'B'C, 

(being  homologous  A  of  similar  polygons). 

Z  ABE  =  Z  A'B'  E', 

(being  homologous  A  of  similar  A  ). 

.'.Z  ABC-  ZABE^Z  A'B'C  ~Z  A'B' E'. 
That  is  Z  EBC  =  ZE'B'C. 


SIMILAR    POLYGONS. 


157 


Now 


also 


EB         AB 

WB'  ~  A'B'' 
(being  homologous  sides  of  similar  & ) ; 

BC   =   AB 

WC'        A'B1' 
{being  homologous  sides  of  similar  polygons). 

EB  BC 


E'  B' 


B'  C" ' 


Ax.  1 

§284 


. ' .  A  E  B  C  and  E'  B'  C  are  similar, 

(having  an  Z  of  the  one  equal  to  an  A  of  the  other,  and  the  including  sides 

'proportional). 

In  like  manner  we  may  prove  AECD  similar  to  A  E' C D'. 

Q.  E.  D. 

Proposition  XVI.     Theorem. 

295.   The  perimeters  of  two  similar  polygons  have  the 
same  ratio  as  any  two  homologous  sides. 


B  C 

Let  the  two  similar  polygons  be  ABODE  and  A'B' CD' E', 
and  let  P  and  P'  represent  their  perimeters. 

We  are  to  prove      P  :  P'  :  :  A  B  :  A'B'. 

AB  :  A'B'  :  :  BC  :  B'  C  :  :  CD  :  C^etc.     §  278 
(the  homologous  sides  of  similar  polygons  are  proportional). 

.'.  AB  +  BC,  etc.   :  A'B'  +  B'C,  etc.   :  :  AB  :  A'B',     §  26G 
(in  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 


That  is 


P  :  P' 


AB  :  A'B' 


Q.  E.  D. 


158 


GEOMETBY. BOOK   III. 


Pboposition  XVII.     Theobem. 

296.   The  homologous  altitudes  of  two  similar  triangles 
have  the  same  ratio  as  any  two  homologous  sides. 


In  the  two   similar  triangles  ABC   and  A'B'C,    let 
the  altitudes  be  BO  and  B'O'. 


We  are  to  prove 


BO 


AB 

A<  B' 


B'O1 

In  the  rt.  A  B  0  A  and  B'  0'  A', 

Z.  A=  Z  A'  §278 

(being  homologous  A  of  the  similar  A  A  B  C  and  A'  B1  C). 

.-.  A  B  0  A  and  A  B'  0'  A'  are  similar,  §  281 

(two  rt.  A  having  an  acute  Z  of  the  one  equal  to  an  acute  Z  of  the  other  are 

similar). 

.'.  their  homologous  sides  give  the  proportion 


BO 


AB 


B'O'        A'B' 


Q.  E.  D 


297.  Cob.  1.    The  homologous  altitudes  of  similar  triangles 
have  the  same  ratio  as  their  homologous  bases. 


SIMILAR    POLYGONS.  159 


§278 


In  the  similar  A  A  B  C  and  A'  B'  C, 

AC         AB 

A7^'  ~  A'B'' 

(the  homologous  sides  of  similar  A  are  proportional). 

And  in  the  similar  A  B  0  A  and  B'  0'  A', 

B0   -   AB  S296 

•    B°         AG  Ax.l 


B'  0'        A>  C ' 

298.  Cor.  2.    The  homologous  altitudes  of  similar  triangles 
have  the  same  ratio  as  their  perimeters. 

Denote  the  perimeter  of  the  first  by  P,  and  that  of  the 
second  by  P'. 

Then  L  =  AA  ,  S  295 

(the  perimeters  of  two  similar  polygons  luivc  the  same  ratio  as  any  two 
homologous  sides). 

But  1°.  =  AJL }  §  296 


Ax.  1 


Ex.  1.  If  any  two  straight  lines  be  cut  by  parallel  lines, 
show  that  the  corresponding  segments  are  proportional. 

2.  If  the  four  sides  of  any  quadrilateral  be  bisected,  show  that 
the  lines  joining  the  points  of  bisection  will  form  a  parallelo- 
gram. 

3.  Two  circles  intersect;  the  line  A  H KB  joining  their 
centres  A,  B,  meets  them  in  //,  K.  On  A  B  is  described  an 
equilateral  triangle  ABC,  whose  sides  B C,  A  C,  intersect  the 
circles  in  F,  E.  F  E  produced  meets  B  A  produced  in  P.  Show 
that  as  PA  is  to  P  K  so  is  C  F  to  CE.  and  so  also  is  PH  to  PB. 


B'O' 

A'B'' 

BO 

P 

Wo1 

"  P' 

160 


GEOMETRY. 


BOOK    III. 


Proposition  XVIII.     Theorem. 

299.  In  any  triangle  the  product  of  two  sides  is  equal 
to  the  product  of  the  segments  of  the  third  side  formed  by  the 
bisector  of  the  opposite  angle  together  with  the  square  of  the 
bisector. 


Let  Z  BA  C  of  the  A  A  B  C  be  bisected  oy  cue  straight 
line  AD. 
We  are  to  prove      BAXAC  =  BDXDC+AD2. 

Describe  the  O  A  B  C  about  the  A  A  B  C ; 
produce  A  D  to  meet  the  circumference  in  E,  and  draw  E  C. 
Then  in  the  A  A  B  D  and  AEG, 

ZBAD  =  ZCAE,  Hyp. 

Z  B  =  Z  E,  §203 

{each  being  measured  by  I  the  arc  AC). 

.\AABDa,if&AEC  are  similar,  §  280 

(two  A  are  similar  when  two  A  of  the  one  are  equal  respectively  to  two  A 
of  tha  other). 

B  A,  the  longest  side  of  the  one, 
:  E  A,  the  longest  side  of  the  other, 
:  A  D,  the  shortest  side  of  the  one, 
:  A  C,  the  shortest  side  of  the  other ; 
BA        AD 

EA  ~  AC' 

(homologous  sides  of  similar  A  are  proportional). 

.\BAXAC  =  EAXAD. 
But  EAX  AD  =  (ED  +  A  D)  A  A 

.'.  BA  X  A  C  =  ED  X  A  D  +  A  D\ 

But  EDXAD  =  BDXDC, 

(the  segments  of  two  chords  in  a  Q  which  intersect  each  other  are 
reciprocally  proportional). 

Substitute  in  the  above  equality  B  D  X  D  C  for  E  D  X  A  D, 
then  BAX  AC  =  BDX  DC  +  AD*. 

Q.  E.  D. 


Whence 


or, 


§278 


§290 


SIMILAR    POLYGONS. 


161 


Proposition  XIX.     Theorem. 

300.  In  any  triangle  the  product  of  two  rides  is  equal  to 
the  product  of  the  diameter  of  the  circumscribed  circle  by  the 
perpendicular  let  fall  upon  the  third  side  from  the  vertex  of 
the  opposite  angle. 


Let  ABC   be  a  triangle,  and  AD   the  perpendiculai 
from  A  to  BC. 

Describe  the  circumference  ABC  about  the  A  A  BC. 

Draw  the  diameter  A  E,  and  draw  E  C. 

We  are  to  prove      BAXAC  =  EAXAD. 

In  the  A  ABD  and  AEC  . 

Z  BDA  is  art.  Z, 


Z  EC  A  is  art,  Z, 
(Jbeing  inscribed  in  a  semicircle). 

.'./.  BDA  =Z  EC  A. 

£B  =  /.E, 

(each  being  measured  by  \  the  arc  A  (T). 
.'.  A  AB  D  and  A  E  C  are  similar, 


Cons. 
§204 

§  203 
§  281 


(two  rt.  A  having  an  acute  Z  of  the  one  equal  to  an  acute  Z.  of  the  other  are 

similar). 


Whence 


or, 


BA,  the  longest  side  of  the  one, 
E  A,  the  longest  side  of  the  other, 
A  D,  the  shortest  side  of  the  one, 
A  C,  the  shortest  side  of  the  other ; 

BA    _  AD 

EA~  AC' 

.BAX  AC  =  EAX  AD. 


§  278 


Q.  E.  D. 


162 


GEOMETRY.  —  BOOK  III. 


Proposition  XX.     Theorem. 

301.  The  product  of  the  two  diagonals  of  a  quadrilateral 
inscribed  in  a  circle  is  equal  to  the  sum  of  the  products  of  its 
opposite  sides. 


Let  ABC  B  be  any  quadrilateral  inscribed  in  a  circle, 
AC  and  BB  its  diagonals. 

We  are  to  prove      BDXAC  =  ABXCD  +  ADXBC. 

Construct  Z  ABE  =  Z  BBC, 

and  add  to  each  Z  E BD. 

Then  in  the  A  ABB  and  B C E, 

ZABB  =  ZCBEf  Ax.  2 

and  ZBBA=ZBCE,  §203 

(each  being  measured  by  |  the  arc  A  B). 

.'.  A  A  B  D  and  B  C E,  are  similar,  §  280 

(two  A  are  similar  when  two  A  of  the  one  are  equal  respectively  to  two  A 
of  the  other). 

Whence       A  B,  the  medium  side  of  the  one, 
C  E,  the  medium  side  of  the  other, 
B  B,  the  longest  side  of  the  one, 
B  C,  the  longest  side  of  the  other, 


SIMILAR   POLYGONS.  163 


§278 


AD  _    BD 
CE  ~  BC' 
(the  homologoics  sides  of  similar  A  are  proportional). 

.'.BD  X  CE  =  AD  X  BC. 

Again,  in  the  A  A  B  E  and  B  C  D, 

Z  ABE  =  Z  DBC,  Cons. 

and  ZBAE  =  ZBDC,  §203 

(each  being  measured  by  $  of  the  arc  B  C). 

.'.  A  A  B  E  and  B  CD  are  similar,  §  280 

(two  A  are  similar  when  two  A  of  the  one  are  equal  respectively  to  two  A 
of  the  other). 

"Whence       A  B,  the  longest  side  of  the  one, 
B  D,  the  longest  side  of  the  other, 
A  E,  the  shortest  side  of  the  one, 
CD,  the  shortest  side  of  the  other. 

or,  ^  =  11,  §278 

BD        CD' 

(the  homologous  sides  of  similar  A  are  proportional). 
.'.BD  X  AE  =  ABX  CD. 
But  BDXCE  =  ADXBC. 

Adding  these  two  equalities, 

BD  (AE+  CE)  =  ABX  CD  +  ADX  BC, 
or         BDXAC  =  ABXCD  +  ADXBC. 

Q.  E.  D. 


Ex.  If  two  circles  are  tangent  internally,  show  that  chords 
of  the  greater,  drawn  from  the  point  of  tangency,  are  divided 
proportionally  by  the  circumference  of  the  less. 


164  GEOMETRY.  —  BOOK   III. 


On  Constructions. 
Proposition  XXI.     Problem. 
302.   To  divide  a  given  straight  line  into  equal  parts. 
A^— . 7b 

Let  A  B  be  the  given  straight  line. 

It  is  required  to  divide  A  B  into  equal  parts. 

From  A  draw  the  indefinite  line  A  0. 

Take  any  convenient  length,  and  apply  it  to  A  0  as  many 
times  as  the  line  A  B  is  to  be  divided  into  parts. 

From  the  last  point  thus  found  on  A  0,  as  C,  draw  C  B. 

Through  the  several  points  of  division  on  A  0  draw  lines 
II  to  CB. 

These  lines  divide  A  B  into  equal  parts,  §  274 

(if  a  series  of  lis  intersecting  any  two  straight  lines,  intercept  equal  parts 
on  one  of  these  lines,  they  intercept  equal  parts  on  the  other  also). 

Q.  E.  F. 


Ex.   To  draw  a  common  tangent  to  two  given  circles. 

I.  When  the  common  tangent  is  exterior. 

II.  When  the  common  tangent  is  interior. 


CONSTRUCTIONS.  165 


Proposition  XXII.     Problem. 

303.  To  divide  a  given  straight  line  into  parts  pro- 
portional  to  any  number  of  given  lines. 

H  K  B 

■A-*z^ v — " r , 


\          \ 

n 

0 

-X 

Let  A  B,  m,  n,  and  o  be  given  straight  lines. 

It  is  required  to  divide  A  B  into  parts  proportional  to  the 
given  lines  m,  n,  and  o. 

Draw  the  indefinite  line  A  X. 

On  A  X  take  A  C  =  m, 

CE  =  n, 

and  EF=o. 

Draw  FB.     From  E  and  C  draw  E  K  and  C  H  II  to  F  B. 
K  and  //  are  the  division  points  required. 

For  f4£V.^.«^»  §275 

\AE)        AC        CE         EF 

(i  line  drawn  through  two  sides  of  a  A  II  to  tlte  third  side  divides  tliosr 
sides  proportional! >t). 

.'.AH  :  HK  :  KB  :  :  AC  :  CE  :  E F. 
Substitute  nt,  n,  and  o  for  their  equals  AC,  C  E,  and  E  F. 
Then  A  H  :  HK  :  KB  : :  m  :  »  :  o. 

Q.  E.  F 


166  GEOMETRY. BOOK   III. 


Proposition  XXIII.     Problem. 

304.    To  find  a  fourth    proportional  to   three    given 
straight  lines. 

B  F  m 


S    '-B 


Let  the  three  given  lines  be  m,  n,  and  o. 

It  is  required  to  find  a  fourth  proportional  to  m,  n,  and  o. 

Take  A  B  equal  to  n. 

Draw  the  indefinite  line  A  R,  making  any  convenient  Z 
with  A  B. 

On  A  R  take  A  C  =  m,  and  0  S  =  o. 

Draw  CB. 

From  S  draw  JSFW  to  C  B,  to  meet  A  B  produced  at  F. 

B  F  is  the  fourth  proportional  required. 

For,  AG  :  AB  :  :  OS  :  B F,  §  275 

(a  line  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those  sides 
proportionally). 

Substitute  on,  n,  and  o  for  their  equals  AC,  AB,  and  G  S. 

Then  m  :  n  :  :  o  :  B  F. 

Q.  E.  F. 


CONSTRUCTIONS.  167 


Proposition  XXIV.     Problem. 

305.  To  find  a  third  proportional  to  two  given  straight 

lines. 

A 


A B 

A C 


Let  A  B  and  A  G  be  the  two  given  straight  lines. 

It  is  required  to  find  a  third  proportional  to  A  B  and  A  G. 

Place  A  B  and  A  G  so  as  to  contain  any  convenient  A. 

Produce  A  B  to  D,  making  BD  =  AG. 

Join  BO. 

Through  D  draw  D  E  II  to  B  G  to  meet  A  0  produced  at  E. 

C  E  is  a  third  proportional  to  A  B  and  AG.         §  251 

£5- £S«  §275 

(a  line  drawn  throiigh  two  sides  of  a  A  II  to  the  third  side  divides  those  sides 
proportionally). 

Substitute,  in  the  above  equality,  A  C  for  its  equal  B  D ; 

Then  d^  =  ^, 

AG        GE' 

or,  A  B  :  A  G  :  :  A  G  :  CE. 

Q.  E.  F. 


168  GEOMETRY.  —  BOOK   III. 

Proposition  XXV.     Problem. 

306.   To  find  a  mean  proportional  between   two  given 
lines. 


B * 

Let  the  two  given  lines  be  m  and  n. 
It  is  required  to  find  a  mean  proportional  between  m  and  n. 
On  the  straight  line  A  E 

take  AG  =  m,  and  G B  =  n. 

On  A  B  as  a  diameter  describe  a  semi-circumference. 

At  G  erect  the  _L  G  H. 

G  H  is  a  mean  proportional  between  m  and  n. 

Draw  II B  and  HA. 

The  Z  A  HB  is  a  rt.  Z,  §  204 

(being  inscribed  in  a  semicircle), 

and  HG  is  a  J_  let  fall  from  the  vertex  of  a  rt.  Z  to  the 

hypotenuse. 

.'.AG  :  GH  ::  GH  :  G B,  §289 

<7/*e  _L  let  fall  from  the  vertex  of  the  rt.  Z.  to  the  hypotenuse  is  a  mean  pro- 
portional between  the  segments  of  the  hypotenuse). 

Substitute  for  A  G  and  G  B  their  equals  m  and  n. 

Then  m  :  G H  :  :  GH  :  n.  Q  E  F 

307.  Corollary.  If  from  a  point  in  the  circumference  a 
perpendicular  be  drawn  to  the  diameter,  and  chords  from  the  point 
to  the  extremities  of  the  diameter,  the  perpendicidar  is  a  mean  pro- 
portional between  the  segments  of  the  diameter,  and  each  chord  is  a 
mean  proportional  between  its  adjacent  segment  and  the  diameter. 


CONSTRUCTIONS.  169 


Proposition  XXVI.     Problem. 

308.   To  divide  one  side  of  a  triangle  into  two  parts 
proportional  to  the  other  two  sides. 


B  E 

Let  ABC  be  the  triangle. 

It  is  required  to  divide  the  side  B  C  into  two  such  parts  that 
the  ratio  of  these  two  parts  shall  equal  the  ratio  of  the  other  two 
sides,  A  C  and  A  B. 

Produce  C  A  to  F,  making  A  F  =  A  B. 
Draw  FB. 
From  A  draw  A  E  II  to  FB. 

E  is  the  division  point  required. 

For  9A.  =  9JL.  §  275 

AF        EB  S 

(a  line  drawn  through  two  sides  of  a  AW  to  the  third  side  divides  those  sides 
proportionally). 

Substitute  for  A  F  its  equal  A  B. 

Then  £A  =  C*. 

AB       EB 

Q.  E.  F. 

309.  Corollary.   The  line  A  E  bisects  the  angle  CAB. 
For  /1F=ZABF,  §112 

(being  opposite  equal  sides). 

ZF=ZCAF,  §70 

(being  ext.-int.  A ). 

ZAJ3F=ZBAFJ,  §68 

(being  alt.-int.  A ). 

.'.ZCAE=ZBAE.  Ax.  1 

310.  Def.  A  straight  line  is  said  to  be  divided  in  extreme 
and  mean  ratio,  when  the  whole  line  is  to  the  greater  segment 
as  the  greater  segment  is  to  the  less. 


170 


GEOMETRY. BOOK   III. 


Proposition  XXVII.     Problem. 
311.  To  divide  a  given  line  in  extreme  and  mean  ratio. 


S 


/ 


H  B 

Let  AB  be  the  given  line. 

It  is  required  to  divide  A  B  in  extreme  and  mean  ratio. 

At  B  erect  a  J_  B  G,  equal  to  one-half  of  A  B. 

From  G  as  a  centre,  with  a  radius  equal  to  G  B,  describe  a  O. 

Since  A  B  is  J_  to  the  radius  GB  at  its  extremity,  it  is 
tangent  to  the  circle. 

Through  G  draw  A  J),  meeting  the  circumference  in  E  and  D. 
OnAB  take  AH  =  AE. 
H  is  the  division  point  of  A  B  required. 

For  AD  :  AB  ::  AB  :  AE,  §  292 

{if  from  a  point  without  the  circumference  a  secant  and  a  tangent  be  drawn, 
the  tangent  is  a  mean  proportional  between  the  whole  secant  and  the  part 
without  the  circumference). 


Then     AD-  AB  -  AB  :  :  A  B  -  A  E  :  A  E. 


265 


CONSTRUCTIONS.  171 


Since  A  B  —  2  G  B,  Cons. 

and  ED  =  2  GB, 

(the  diameter  of  aO  being  'twice  the  radius), 

AB  =  ED.  Ax.  1 

.'.AD-AB  =  AD-ED  =  AE. 

But  AE  =  AH,  Cons. 

.'.  A  D  -  A  B  =  A  H.  Ax.  1 

Also         AB-AE  =  AB-AH  =  HB. 

Substitute  these  equivalents  in  the  last  proportion. 

Then  AH  :  AB  :  :  HB  :  AH. 

Whence,  by  inversion,  AB  :  AH  :  :  AH  :  HB.      §  263 

.'.  A  B  is  divided  at  H  in  extreme  and  mean  ratio. 

Q.  E.  F. 

Eemark.  A  B  is  said  to  be  divided  at  H,  internally,  in 
extreme  and  mean  ratio.  If  BA  be  produced  to  H',  making 
A  H'  equal  to  A  D,  A  B  is  said  to  be  divided  at  H',  externally, 
in  extreme  and  mean  ratio. 

Prove  AB  :  AH'  :  :  AH  :  W B. 

When  a  line  is  divided  internally  and  externally  in  th3 
same  ratio,  it  is  said  to  be  divided  harmonically. 

Thus^5  ± £__£ £?  is  divided  harmoni- 
cally at  G  and  D,  if  C  A  :GB::DA:DB;  that  is,  if  the  ratio 
of  the  distances  of  G  from  A  and  B  is  equal  to  the  ratio  of  the 
distances  of  D  from  A  and  B. 

This  proportion  taken  by  alternation  gives  : 

AG  :AD::BG:BD;  that  is,  G D  is  divided  harmoni- 
cally at  the  points  B  and  A.  The  four  points  A,  B,  C,  D,  are 
called  harmonic  points ;  and  the  two  pairs  A,  B,  and  G,  D,  are 
called  conjugate  points. 


Ex.  1.    To  divide  a  given  line  harmonically  in  a  given  ratio. 
2.    To  find  the  locus  of  all  the  points  whose  distances  from 
two  given  points  are  in  a  given  ratio. 


/  / 

\  \ 

\ 

\       \ 

\ 

\ 

\  / 

\ 

\  / 

\    / 
\/ 

B' 

a 

172  GEOMETRY. BOOK   III. 


Proposition  XXVIII.     Problem. 

312.  Upon  a  given  line  homologous  to  a  given  side  of  a 
given  polygon,  to  construct  a  polygon  similar  to  the  given 
polygon. 

E 


I 
I 


Let  A'  E  be  the  given  line,  homologous  to  A  E  of  the 
given  polygon  ABC D E. 

It  is  required  to  construct  on  A1  E'  a  polygon  similar  to  the 
given  polygon. 

From  E  draw  the  diagonals  E B  and  EG. 

From  E'  draw  E>  B',  making  Z  A'  E'  B'  =  Z  A  E  B. 

Also  from  A1  draw  A'  B',  making  Z  B'  A'  E'  =  Z  B  A  Ey 

and  meeting  E'  B'  at  B'. 

The  two  A  A  B  E  and  A1  B'  E'  are  similar,  §  280 

(two  A  are  similar  if  they  have  two  A  of  the  one  equal  respectively  to  two  A 
of  the  other). 

Also  from  E'  draw  E'  C",  making  Z  B'  E'  C  =  Z  B  E  C. 

From  B'  draw  B'  C,  making  Z  E'  B> '  C '  =  Z  E  B  C, 

and  meeting  E'  C  at  C. 

Then  the  two  A  EB  G  and  E'  B'  G'  are  similar,      §  280 
(two  &  are  similar  if  they  have  two  A  of  the  one  equal  respectively  to  two  A 
of  the  other). 

In  like  manner  construct  A  E'  G'  B'  similar  to  A  E  G  D. 

Then  the  two  polygons  are  similar,  §  293 

(two  polygons  composed  of  the  same  member  of  A  similar  to  each  other  and 
similarly  placed,  are  similar). 

.'.  A'  B'  G'  D'  E'  is  the  required  polygon. 

Q.  E.  F. 


EXEfcClSES.  173 


Exercises. 

1.  A  B  C  is  a  triangle  inscribed  in  a  circle,  and  B  D  is  drawn 
to  meet  the  tangent  to  the  circle  at  A  in  D,  at  an  angle  ABB 
equal  to  the  angle  ABC;  show  that  A  C  is  a  fourth  propor- 
tional to  the  lines  B  D,  A  D,  A  B. 

2.  Show  that  either  of  the  sides  of  an  isosceles  triangle  is  a 
mean  proportional  between  the  base  and  the  half  of  the  segment 
of  the  base,  produced  if  necessary,  which  is  cut  off  by  a  straight 
line  drawn  from  the  vertex  at  right  angles  to  the  equal  side. 

3.  A  B  is  the  diameter  of  a  circle,  D  any  point  in  the  circum- 
ference, and  G  the  middle  point  of  the  arc  AD.  If  A  C,  A  D, 
B  C  be  joined  and  A  D  cut  B  C  in  E,  show  that  the  circle  cir- 
cumscribed about  the  triangle  A  E  B  will  touch  A  C  and  its 
diameter  will  be  a  third  proportional  to  B  C  and  A  B. 

4.  From  the  obtuse  angle  of  a  triangle  draw  a  line  to  the 
base,  which  shall  be  a  mean  proportional  between  the  segments 
into  which  it  divides  the  base. 

5.  Find  the  point  in  the  base  produced  of  a  right  triangle, 
from  which  the  line  drawn  to  the  angle  opposite  to  the  base 
shall  have  the  same  ratio  to  the  base  produced  which  the  per- 
pendicular has  to  the  base  itself. 

6.  A  line  touching  two  circles  cuts  another  line  joining  their 
centres ;  show  that  the  segments  of  the  latter  will  be  to  each 
other  as  the  diameters  of  the  circles. 

7.  Required  the  locus  of  the  middle  points  of  all  the  chords 
of  a  circle  which  pass  through  a  fixed  point. 

8.  0  is  a  fixed  point  from  which  any  straight  line  is  drawn 
meeting  a  fixed  straight  line  at  P ;  in  0  P  a  point  Q  is  taken 
such  that  0  Q  is  to  0  P  in  a  fixed  ratio.  Determine  the  locus 
of  Q. 

9.  0  is  a  fixed  point  from  which  any  straight  line  is  drawn 
meeting  the  circumference  of  a  fixed  circle  at  P  ;  in  0  P  a  point 
Q  is  taken  such  that  0  Q  is  to  0  P  in  a  fixed  ratio.  Determine 
the  locus  of  Q. 


BOOK  IV. 


COMPARISON    AND    MEASUREMENT    OF    THE    SUR- 
FACES  OF  POLYGONS. 

Proposition  I.     Theorem. 

313.   Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 


D 


D 


0  "      ~         O 

Let  the  two  rectangles  be  AC  and  A  F,  having  the 
the  same  altitude  A  D. 

rect.  A  0  _    AB 

iecLAF~~  AE' 


We  are  to  prove 


Then 


Case  I.  —  When  A  B  and  A  E  are  commensttrable. 
Find  a  common  divisor  of  the  bases  A  B  and  A  E,  as  A  0. 
Suppose  A  0  to  be  contained  in  A  B  seven  times  and  in 
A  E  four  times. 

AB  =    7 
AE  ~  4' 

At  the  several  points  of  division  on  A  B  and  A  E  erect  Js . 
The  rect.  A  C  will  be  divided  into  seven  rectangles, 
and  rect.  A  F  will  be  divided  into  four  rectangles. 

These  rectangles  are  all  equal,  for  they  may  be  applied  to 
each  other  and  will  coincide  throughout. 


But 


rect  A  G 

7 

rect  A  F 

4 

AB 

7 

AE 

4 

rect  A  G 
rect  A  F 

AB 
AE 

COMPARISON    AND    MEASUREMENT    OF    POLYGONS. 


175 


CASE  II.  —  When  A  B  and  A  E  are  incommensurable. 


D 


D 


II 


B 


K 


E 


Divide  A  B  into  any  number  of  equal  parts,  and  apply  one 
of  these  parts  to  A  E  as  often  as  it  will  be  contained  in  A  E. 

Since  A  B  and  A  E  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  i  to  a  point  K,  leaving  a  re- 
mainder K  E  less  than  one  of  these  parts. 

Draw  JSTJEMI  to  E F. 

Since  A  B  and  A  K  are  commensurable, 

rect.AH  =  AK  Case  j 

rect.  AC  ~    AB 

Suppose  the  number  of  parts  into  which  A  B  is  divided  to 
be  continually  increased,  the  length  of  each  part  will  become  less 
and  less,  and  the  point  K  will  approach  nearer  and  nearer  to  E. 

The  limit  of  A  it  will  be  A  E,  and  the  limit  of  rect.  A  H 
will  be  rect.  A  F. 

.'.the  limit  of  —  will  be  UH, 
AB  AB 

j  i.u    v    ■*.    c  reck  AH     .11  •     rect.  A  F 
and  the  limit  ot will  be 


rect.  A  C 


rect.  A  C 


Now  the  variables   and  . are   always   equal 

A  B  rect.  AC  J       H 


however  near  they  approach  their  limits  ; 

rect.  A  F 


.'.  their  limits  are  equal,  namely, 


AE 


TGct.AC       AB 


§199 


Q.  E.  D. 

314.  Corollary.  Two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes.  By  considering  the  bases  of 
these  two  rectangles  A  D  and  A  D,  the  altitudes  will  be  A  B  and 
A  E.  But  we  have  just  shown  that  these  two  rectangles  are  to 
each  other  as  A  B  is  to  A  E.  Hence  two  rectangles,  with  the 
same  base,  or  equal  bases,  are  to  each  other  as  their  altitudes. 


176 


GEOMETRY.  —  BOOK    IV. 


Another  Demonstration. 
Let  A  C  and  A1  C  be  two  rectangles  of  equal  altitudes. 
P  C  O  Pi 


F       E       D       A  A'  D<  E>  F>  G' 

rect.  AC         AD 


We  are  to  prove 


rect.  A'C        A' J)' 


Let  b  and  &',  S  and  S'  stand  for  the  bases  and  areas  of  these 
rectangles  respectively. 

Prolong  A  D  and  A'  D\ 

Take  AD,  D E,  E F  .  .  .  .  m  in  number  and  all  equal, 

and  A'  D',  D'  E',  E'  F',  F' G'  .  .  .  .  n  in  number  and  all  equal. 

Complete  the  rectangles  as  in  the  figure. 

Then  base  AF  =  mb, 

and  base  A' ' G' '  ==  nb' ; 

rect.  A  P  =  mS, 
and  rect.^/P/=^^/. 

Now  we  can  prove  by  superposition,  that  if  A  F  be  >  A'  G', 
rect.  A  P  will  be  >  rect.  A'  P' ;  and  if  equal,  equal ;  and  if  less, 
less. 

That  is,  if  mb  be  >  nb',  m S  is  >  n S' ;  and  if  equal, 
equal ;  and  if  less,  less. 

Hence,  b  :  b'  :  :  S  :  &,        Euclid's  Def.,  §  272 

Q.  E.  D. 


COMPARISON   AND   MEASUREMENT    OF    POLYGONS. 


177 


Proposition  II.     Theorem. 

315.   Two  rectangles  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 


a' 


Rf 


J 


b  b'  b 

Let  R  and  R'  be  two  rectangles,  having  for  their  bases 
b  and  b',  and  for  their  altitudes  a  and  a'. 
R_         aXb 
R'  ~ 


We  are  to  prove 


§314 


§  313 


a'  X  y 

Construct  the  rectangle  S,  with  its  base  the  same  as  that 
of  R  and  its  altitude  the  same  as  that  of  R'. 

(rectangles  having  the  same  base  are  to  each  other  as  their  altitudes) ; 

and  3**' 

(rectangles  having  the  same  altitude  are  to  each  other  as  their  bases). 
By  multiplying  these  two  equalities  together 
R  aX  b 

R'  ~~  a'  Xbr 

Q.  E.  D. 

316.  Def.    The  Area  of  a  surface  is  the  ratio  of  that  surface 
to  another  surface  assumed  as  the  unit  of  measure. 

317.  Def.    The  Unit  of  measure  (except  the  acre)  is  a  square 
a  side  of  which  is  some  linear  unit ;  as  a  square  inch,  etc. 

318.  Def.    Equivalent  figures  are  figures  which  have  equal 
areas. 

Rem.  In   comparing   the   areas   of  equivalent   figures   the 
symbol  ( = )  is  to  be  read  "  equal  in  area." 


178 


GEOMETRY. BOOK   IV. 


Proposition  III.     Theorem. 

319.  The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 


b  1 

Let  R  be  the  rectangle,  b  the  base,  and  a  the  alti- 
tude ;  and  let  U  be  a  square  whose  side  is  the 
linear  unit. 

We  are  to  prove      the  area  of  R  =  a  X  b. 

R     _  aXb 
U     ,  1X1* 

{two  rectangles  are  to  each  other  as  the  product  of  their  bases  and  altitudes). 

R 


§315 


But 


U 


is  the  area  of  R, 


§316 


the  area  of  R  =  a  X  b. 


Q.  E.  D. 


320.  Scholium.  When  the  base  and  altitude  are  exactly- 
divisible  by  the  linear  unit,  this  proposition  is  rendered  evident 
by  dividing  the  figure  into  squares,  each  equal  to  the  unit  of 


measure.  Thus,  if  the  base  contain  seven  linear  units,  and  the 
altitude  four,  the  figure  may  be  divided  into  twenty-eight 
squares,  each  equal  to  the  unit  of  measure;  and  the  area  of 
the  figure  equals  7X4, 


COMPARISON    AND    MEASUREMENT    OF   POLYGONS.         179 

Proposition  IV.     Theorem. 

321.   The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  and  altitude. 


BE  C     F        B  C    E  F 


A  D 

Let  A  E FD  be  a  parallelogram,  A  D  its  base,  and  G  D 
its  altitude. 

We  are  to  prove     the  area  of  the  EJ  A  E  F  D  =  A  D  X  G D. 

From  A  draw  A  B  II  to  D  G  to  meet  F  E  produced. 

Then  the  figure  A  BG D  will  be  a  rectangle,  with  the  same 
base  and  altitude  as  the  O  A  E  F  D. 

In  the  rt.  A  A  B  E  and  CD  F, 

AB  =  GD,  §126 

(being  opposite  sides  of  a  rectangle). 

and  AE  =  DF,  §134 

(being  opposite  sides  of  a  CD)  ; 

.'.AABE  =  AGDF,  §109 

(two  rt.  A  are  equal,  when  the  hypotenuse  and  a  side  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  a  side  of  the  other). 

Take  away  the  A  G D F and  we  have  left  the  rect.  ABG D. 

Take  away  the  A  A  B  E  and  we  have  left  the  O  A  E  F  D. 

.'.  rect.  ABG  D  =  O  A  EFD.  Ax.  3 

But  the  area  of  the  rect.  ABC  D  =  AD  X  CD,     §  319 

(the  area  of  a  rectangle  equals  the  product  of  its  base  and  altitude). 

.'.  the  area  of  the  O  A  EFD  =  A  D  X  C D.         Ax.  1 

Q.  E.  D. 

322.  Corollary  1.  Parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent. 

323.  Cor.  2.  Parallelograms  having  equal  bases  are  to 
each  other  as  their  altitudes ;  parallelograms  having  equal  alti- 
tudes are  to  each  other  as  their  bases ;  and  any  two  parallelo- 
grams are  to  each  other  as  the  products  of  their  bases  by  their 
altitudes. 


180  GEOMETRY. BOOK    IV. 


Proposition  V.     Theorem. 

324.  The  area  of  a  triangle  is  equal  to  one-half  of  the 
product  of  its  base  by  its  altitude. 


A  B      D 

Let  ABC  be   a    triangle,  AB  its   base,    and  CD  its 
altitude. 

We  are  to  prove      the  area  of  the  A  A  B  C  =  J  A  B  X  CD. 

From  Cdraw  C  ff  \\  to  A  B. 

From  A  draw  A  H  \\  to  B  C. 

The  figure  A  B  C  H  is  a  parallelogram,  §  136 

{having  its  opposite  sides  parallel), 

and  A  C  is  its  diagonal. 

.-.A  ABC  =  A  AHC,  §  133 

(the  diagonal  ofaO  divides  it  into  two  equal  A  ). 

The  area  of  the  ED  ABC H  is  equal  to  the  product  of  its 
base  by  its  altitude.  §  321 

.'.the  area  of  one-half  the  O,  or  the  A  A  B  C,  is  equal  to 
one-half  the  product  of  its  base  by  its  altitude, 

or,  IABXCD. 

2  Q.  E.  D. 

325.  Corollary  1.  Triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

326.  Cor.  2.  Triangles  having  equal  bases  are  to  each  other 
as  their  altitudes ;  triangles  having  equal  altitudes  are  to  each 
other  as  their  bases ;  any  two  triangles  are  to  each  other  as  the 
product  of  their  bases  by  their  altitudes. 


COMPARISON    AND    MEASUREMENT   OF    POLYGONS.        181 


Proposition  VI.     Theorem. 

327.   The  area  of  a  trapezoid  is  equal  to  one-half  the 
sum  of  the  parallel  sides  multiplied  by  the  altitude. 
H  E  C 


A  F  B 

Let  A  B  G II  be  a  trapezoid,  and  EF  the  altitude. 
We  are  to  prove      area  of  A  B  G  H '  =  \  {EG  +  A  B)  E F. 
Draw  the  diagonal  A  C. 
Then  the  area  of  the  A  A  HG  =  \  HG  X  EF,      §  324 

(the  area  of  a  A  is  equal  to  one-half  of  the  'product  of  its  base  by  its  altitude), 

and  the  area  of  the  A  A  B  C  =  J  A  B  X  EF,        §  324 
.-.AAHC+  AABG, 
or,  area  ofABGH=i(HG+AB)  EF. 

Q.  E    D. 

328.  Corollary.  The  area  of  a  trapezoid  is  equal  to  the 
product  of  the  line  joining  the  middle  points  of  the  non-parallel 
sides  multiplied  by  the  altitude ;  for  the  line  0  P,  joining  the 
middle  points  of  the  non-parallel  sides,  is  equal  to  \  (HG 
+  AB).  §142 

.'.by  substituting  0  P  for  %(H  G  +  A  B),  we  have, 
the  area  of  A  B  G H  =  OPX  E F. 

329.  Scholium.  The  area 
of  an  irregular  polygon  may  be 
found  by  dividing  the  polygon 
into  triangles,  and  by  finding 
the  area  of  each  of  these  tri- 
angles separately.  But  the 
method  generally  employed  in 
practice  is  to  draw  the  longest 

diagonal,  and  to  let  fall  perpendiculars  upon  this  diagonal  from 
the  other  angular  points  of  the  polygon. 

The  polygon  is  thus  divided  into  figures  which  are  right 
triangles,  rectangles,  or  trapezoids  ;  and  the  areas  of  each  of  these 
figures  may  be  readily  found. 


182  GEOMETRY.  —  BOOK   IV. 


Proposition  VII.     Theorem. 

330.  The  area  of  a  circumscribed  poly goyi  is  equal  to  one- 
half  the  product  of  the  perimeter  by  the  radius  of  the  in- 
scribed circle. 

B 


Let  ABSQ,  etc.,  be  a  circumscribed  polygon,  and  G 
the  centre  of  the  inscribed  circle. 

Denote  the  perimeter  of  the  polygon  by  P,  and  the  radius 
of  the  inscribed  circle  by  R. 

We  are  to  prove 

ihe  area  of  the  circumscribed  polygon  =  \  P  X  R. 

Draw  G A,  G B,  OS,  etc.; 

also  draw  0  0,  G  D,  etc.,  _L  to  A  B,  B  S,  etc. 

The  area  of  the  A  CA  B  =  \A  B  X  C  0,  §  324 

{the  area  of  a  A  is  equal  to  one-half  the  product  of  its  base  and  altitude). 

The  area  of  the  A  CBS  =  \  B  S  X  CD,  §  324 

.*.  the  area  of  the  sum  of  all  the  A  C  A  B,  CBS,  etc., 
=  i(AB  +  BS,  etc.)  GO,  §  187 

(for  0  0,  CD,  etc.,  are  equal,  being  radii  of  the  same  O). 

Substitute  for  A  B  +  BS  +  SQ,  etc.,  P,  and  for  G  0,  R ; 

then  the  area  of  the  circumscribed  polygon  =  |PX  R. 

Q.  E.  D. 


COMPARISON    AND   MEASUREMENT   OP   POLYGONS. 


183 


Proposition  VIII.     Theorem. 

331.  The  sum  of  the  squares  described  on  the  two  sides 
of  a  right  triangle  is  equivalent  to  the  square  described  on  the 
hypotenuse. 


Let  ABC  be  a  right  triangle  with  its  right  angle  atC. 
We  are  to  prove      AC2  +  CB2  =  AB2 
Draw  0  0  JL  to  A  B. 

Then  AC2  =  AOXAB,  §  289 

{the  square  on  a  side  of  a  rt.  A  is  equal  to  the  product  of  the  hypotenuse  by 
the  adjacent  segment  made  by  Oic  _L  let  fall  from  the  vertex  of  the  rt.  Z)  ; 


and  fit'2  =  BOX  AB, 

By  adding,  AT?  +  F7?=  (A  0  +  B  0)  A  B, 
=  ABX  AB, 

332.  Corollary.    The  side  and  diagonal  A 
of  a  square  are  incommensurable. 

Let  ABGD  be  a  square,  and  AC  the 
diagonal. 

Then  AB2  +  FV2  =  A~C\ 

or,  2  AB2  =  ATC2.  B 

Divide  both  sides  of  the  equation  by  AB2, 

AB2 


§  289 


Q.  E.  D. 


Extract  the  square  root  of  both  sides  the  equation, 
then 


AC          ,_ 


AB 


=  s/Y. 


Since  the  square  root  of  2  is  a  number  which  cannot  be 
exactly  found,  it  follows  that  the  diagonal  and  side  of  a  square 
nro  two  inoouimpnsurable  linos. 


184 


GEOMETRY. BOOK    IV. 


Another  Demonstration. 

833.  The  square  described  on  the  hypotenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  on  the  other 
two 


^y^           /  n 

^  \ 

/ 

\ 

/ 

/ 

\L 

1 

D  L    E 

Let  ABC  be  a  light  A,  having  the  right  angle  BAG. 

We  are  to  prove      BG2  =  BA   +  AG  . 

OnB  G,  GA,  A  B  construct  the  squares  B  E,  CH,  A  F. 

Through  A  draw  A  L  II  to  C  E. 

Draw  A  D  and  FG. 

Z  BACissirt.  Z, 

Z  BAGi&nrt.  Z, 

.  •.  G  A  G  is  a  straight  line. 

Z  CAHis&Tt.  Z, 
.'.  B  A  H  is  a  straight  line. 


and 


Also 


Hyp. 

Cons. 

Cons. 


Now 


Z  DBC  =  Z  FBA, 

(each  being  art.  £). 


Cons. 


COMPARISON   AND    MEASUREMENT    OF    POLYGONS.         185 

Add  to  each  the  A  A  B  G ; 
then  ZABD  =  ZFBG, 

.\AABB  =  AFBC.  §  106 

Now  O  B  L  is  double  A  A  B  D, 

(being  on  the  same  base  B  D,  and  between  the  same  lis,  A  L  and  BD), 

and  square  A  F  is  double  A  F  B  C, 
(being  on  the  same  base  FB,  and  between  the  same  lis,  FB  and  GO); 

.-.  O  BL  =  square  A  F. 

In  like  manner,  by  joining  A  E  and  BK,  it  may  be  proved 
that 

O  CL  =  square  G  H. 

Now  the  square  onBG  =  0  BL  +  O  G L, 

=  square  A  F  +  square  G  Ht 

.-.  BG*  =  FT  +  AG\ 

Q.  E.  D. 


On  Projection. 
334.  Def.    The  Projection  of  a  Point  upon  a  straight  line 
of  indefinite  length  is  the  foot  of  the  perpendicular  let  fall  from 
the  point  upon  the  line.     Thus,  the  projection  of  the  point  G 
upon  the  line  A  B  is  the  point  P. 

C  C 


P  R *         n^P  DD 

Fig.  1.  Fig.  2. 

The  Projection  of  a  Finite  Straight  Line*  as  G  D  (Fig.  1), 
upon  a  straight  line  of  indefinite  length,  as  A  B,  is  the  part  of 
the  line  A  B  intercepted  between  the  perpendiculars  G  P  and 
D  B,  let  fall  from  the  extremities  of  the  line  G  D. 

Thus  the  projection  of  the  line  G  D  upon  the  line  A  B  is 
the  line  P  R. 

If  one  extremity  of  the  line  G  D  (Fig.  2)  be  in  the  line 
A  B,  the  projection  of  the  line  G  D  upon  the  line  A  B  is  the 
part  of  the  line  A  B  between  the  point  D  and  the  foot  of  the 
perpendicular  G  P ;  that  is,  D  P. 


186 


GEOMETRY. BOOK  IV. 


Proposition  IX.     Theorem. 


335.  In  any  triangle,  the  square  on  the  side  opposite  an 
acute  angle  is  equivalent  to  the  sum  of  the  squares  of  the  other 
two  sides  diminished  by  twice  the  product  of  one  of  those 
sides  and  the  projection  of  the  other  upon  that  side. 


Let  C  be  an   acute  angle  of  the  triangle  ABC,  and 
D C  the  projection  of  AC  upon  B C. 

We  are  to  prove      U?  =  WD2  +  J~C*  —  2  B  C  X  D  C. 
If  D  fall  upon  the  base  (Fig.  1), 

DB  =  BC-Z>C; 
If  D  fall  upon  the  base  produced  (Fig.  2), 

DB  =  DC~BC. 

In  either  case  B~B2  =  BC*  +  IfC*  -  2 B C  X  D C. 
Add  A  D   to  both  sides  of  the  equality ; 

then,  JHf  +  fiB2  =  FV2  +  ID2  +  UC2  -2BCXDC. 


But 


AD2  +  WB2=  £B2, 


331 


(the  sum  of  the  squares  on  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse) ; 


and 


ad2  +  irc2  =  jrc* 


331 


Substitute  JTB   and  A  C   for  their  equivalents  in  the  above 
equality ; 


then,      AB2  =  FV2  +  J~C2-2BCXDC. 


Q.  E.  D. 


COMPARISON   AND   MEASUREMENT    OF   POLYGONS.         187 


Proposition  X.     Theorem. 

336.  In  any  obtuse  triangle,  the  square  on  the  side 
opposite  the  obtuse  angle  is  equivalent  to  the  sum  of  the 
squares  of  the  other  two  sides  increased  by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other  on  that 
side, 

A 


Let  G  be  the  obtuse  angle  of  the  triangle  ABC,  and 
G D  be  the  projection  of  A  C  upon  BC  produced. 

We  are  to  prove      IB*  =  Blf  +  £Jf  +  2  B  G  X  D  G. 

DB=BC+ DG 

Squaring,  1TB2  =  Blf  +  Blf  +  2  B  G  X  D  G 

Add  A~lf  to  both  sides  of  the  equality ; 

then,  AD2  +  D~B2  =  E~02  +  AD2  +  DC2  +  2BGXDG. 

But  ID2  +  DB2  =  IB2 ,  §  331 

{the  sum  of  the  squares  on  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse) ; 

and  JTD2  +  Blf  =  JTG2.  §  331 

Substitute  A^B  and  J~D  for  their  equivalents  in  the 
above  equality; 

then,      A~B2  =  Blf  +  A~G2  +  2  BC  X  DG. 

Q.  E.  D. 

337.  Definition.  A  Medial  line  of  a  triangle  is  a  straight 
line  drawn  from  any  vertex  of  the  triangle  to  the  middle  point 
of  the  opposite  side. 


188  GEOMETRY. BOOK   IY. 


Proposition  XI.     Theorem. 

338.  In  any  triangle,  if  a  medial  line  be  drawn  from 
the  vertex  to  the  base : 

I.  The  sum  of  the  squares  on  the  two  sides  is  equivalent 
to  twice  the  square  on  half  the  base,  increased  by  twice  the 
square  on  the  medial  line  ; 

IT.  The  difference  of  the  squares  on  the  two  sides  is 
equivalent  to  twice  the  product  of  the  base  by  the  projection 
of  the  medial  line  upon  the  base. 

A 


In  the  triangle  ABC  let  A  M  be  the  medial  line  and 
M  D  the  projection  of  A  M  upon  the  base  B  C. 
Also  let  AB  be  greater  than  A  C. 

We  are  to  prove 

i.  rtf  ¥  nf  =  2  em2  +  2  in*. 

II.    IB*  -  AG1  =2BCX  MD. 
Since  A  B  >  A  C,  the  Z  A  MB  will  be  obtuse  and  the 
Z.  A  M  C  will  be  acute.  §  HQ 

Then      JTB2  =  BM1  +  AM1  +  2  BM  X  M  D,       §336 

{in  any  obtuse  A  the  square  on  the  side  opposite  the  obtuse  Z.  is  equivalent  to 
the  sum  of  the  squares  on  the  other  two  sides  increased  by  twice  the 
product  of  one  of  those  sides  and  the  projection  of  the,  other  on  that  side)  ; 

and      JT7?  =  m)2  +  AM2  -  2  MCX  MD,  §335 

in  any  A  the  square  on  the  side  opposite  an  acute  Z  is  equivalent  to  the  sum 
of  the  squares  on  the  other  two  sides,  diminished  by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other  upon  that  side). 

Add  these  two  equalities,  and  observe  that  B M  —  MC. 
Then        AB2  +  A~C2  =  2  BM2  +  2  A~M2. 
Subtract  the  second  equality  from  the  first. 
Then  AJ?-A~C2=2BCXMD. 

Q.  E.  D. 


COMPARISON   AND    MEASUREMENT    OF    POLYGONS.         189 

Proposition  XII.     Theorem. 
339.  The  sum  of  the  squares  on  the  four  sides  of  any 
quadrilateral  is  equivalent  to  the  sum  of  the  squares  on  the 
diagonals    together  with  four  times  the  square  of  the  line 
joining  the  middle  points  of  the  diagonals. 
A 


In  the  quadrilateral  A  BCD,  let  the  diagonals  be  A  C 
and   B  D,    and   F  E   the  line  joining  the   middle 
points  of  the  diagonals. 
We  are  to  prove 

JTff  +  BJf  +  (TT)2  +  DA2  =  AC2  +  Bl?  +  4  El*' 
Draw  BE  and  D  E. 

Now       iO2  +  BC1  =  2  (— Y  +  2  rf,  §  338 

(the  sum.  of  the  squares  on  the  two  sides  of  a  A  is  equivalent  to  twice  the  square 
on  half  t/ie  base  increased  by  twice  the  square  on  the  medial  line  to  the  base), 

and         CI?  +  m2  =  2  (A-£Y  +  2DE2.  §  338 

Adding  these  two  equalities, 

.O2  +  BO2  +  01?  +  DA2  =  4  (4^V  +  2  (^  +  E~E\ 

But        BE2  +  JTE1  =  2  (^V  +  2 EF2,  §  338 

(the  sum  of  the  squares  on  the  two  sides  of  a  A  is  equivalent  to  twice  the  square 
on  half  the  base  increased  by  twice  the  square  on  t/te  medial  line  to  the  base). 

Substitute  in  the  above  equality  for  (BE2  +  DE2)  its 
equivalent ; 

theni^4-^2  +  Z7Z52+ro2  =  4(^2  +  4(^)2+4^a 

=  IC2  +  BD2  +  4  ET 

Q.  E.  D. 

340.  Corollary.  The  sum  of  the  squares  on  the  four  sides 
of  a  parallelogram  is  equivalent  to  the  sum  of  the  squares  on  the 
diagonals. 


190 


GEOMETRY. 


BOOK    IV. 


Proposition  XIII.     Theorem. 


341.  Two  triangles  having  an  angle  of  the  one  equal  to 
an  angle  of  the  other  are  to  each  other  as  the  products  of  the 
sides  including  the  equal  angles. 


Let  the  triangles  ABC  and  ADE  have  the  common 
angle  A. 


We  are  to  prove 

Draw  BE. 

Now 


AABC        ABXAG 


AADE        ADXAE 


AABC        AC 


AABE       AE 

(A  having  the  same  altitude  are  to  each  other  as  their  bases). 


Also 


AABE 


AB 
AD 


AADE 

(A  leaving  the  same  altitude  are  to  each  other  as  their  bases). 

Multiply  these  equalities ; 


326 


§  326 


then 


AABC  =   ABX  AG 
AADE        ADX  AE 


Q.  E.  D. 


COMPARISON   AND    MEASUREMENT    OF    POLYGONS.         191 

Proposition  XIV.     Theorem. 
342.  Similar  triangles  are  to  each  other  as  the  squares 
on  their  homologous  sides. 


0  ~  0'  B' 

Let  the  two  triangles  be  AG B  and  A'C'B'. 

w         .  AACB  Aff 

We  are  to  prove      = 

A  A'C'B'       AT^ 

Draw  the  perpendiculars  C  0  and  C  O1. 

Then     AACB    =    *BXC0     Z  **  X    «£-,      §  326 

A  A'C'B'        A'B'XC'O'        A' B'        CO'        * 

(two  A  are  to  each  other  as  the  products  of  their  bases  by  their  altitudes). 

But  ±2=™,  §297 

A'B'        CO''  S 

(the  homologous  altitudes  of  similar  &  have  the  same  ratio  as  their  homolo- 
gous bases). 

Substitute,  in  the  above  equality,  for its  equal  ; 

1        J  CO'  l       A'B' 

,,  AACB         AB         AB  _   A^ 

then      =  v 

a    a  i  ni  m         a  i  »/  ^ 


Q.  E.  D. 


192 


GEOMETRY. BOOK   IV. 


Proposition  XV.     Theorem. 

343.   Two   similar  polygons   are   to   each   other  as  the 
squares  on  any  two  homologous  sides. 
B  C 


F  E 

Let    the    two  similar  polygons   be   ABC,    etc..    and 

A' BO,  etc.    . 

m  •  ABC,  etc.  ATE2 

We  are  to  prove    : =  . 

A'  B'  C,  etc.        jrrfi 

From  the  homologous  vertices  A  and  A'  draw  diagonals. 

AB         BC 

A'B'  ~ 


Now 


CD       , 

,  etc., 

B'C        CD' 


(similar  polygons  have  their  homologous  sides  proportional)  ; 


.'.by  squaring, 


CD1 


ATB2         Ftf  _ 

jFB'2       Ftf       CU'2 


,  etc. 


The  AABC,ACD,  etc.,  are  respectively  similar  to  A' B'C, 


A'  C  D\  etc., 


294 


(two  similar  polygons  are  composed  of  the  same  number  of  &  similar  to  each 
other  and  similarly  placed). 


A  ABC 


nf 


A  A' B'C       AHS'2 
(similar  A  are  to  each  other  as  the  squares  on  their  homologous  sides) 

AACD  0~D2 


§  342 


and 


A  A' CD' 


CD'2 


§  34? 


COMPARISON  AND  MEASUREMENT  OF  POLYGONS.    193 

WW1      JJlr 

AABC  AACD 


AA'B'C        A  A'  CD1 

In  like  manner  we  may  prove  that  the  ratio  of  any  two  of 
the  similar  A  is  the  same  as  that  of  any  other  two. 

AABC  AACD  A  APE  AAEF 

'  '  AA'B'C  ""  A  A' C" D'  ~  A  A' D' E'  ™  AA'E'F' 

AABC  +  ACD  +  ADE-V  AEF  AABC 

AA'B'C'  +  A'C'D'  +  A'D'E'  +  A'  E'  F'~  A  A1  B'  C' 

(in  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 

But  AABC   -I*-,  §342 

AA'B'C        jrgi'  S 

(similar  &  are  to  each  other  as  the  squares  on  their  homologous  sides)  ; 

.     the  polygon  ABC,  etc.       _  £1? 
the  polygon  A'  B'  C,  etc.      .  A7H'2  ' 

Q.  E.  D. 

344.  Corollary  1.     Similar  polygons  are  to  each  other  as 
the  squares  on  any  two  homologous  lines. 

345.  Cor.  2.    The  homologous  sides  of  two  similar  poly- 
gons have  the  same  ratio  as  the  square  roots  of  their  areas. 

Let  S  and  S'  represent  the  areas  of  the  two  similar  polygons 
A  B  G,  etc.,  and  A'  B'  C,  etc.,  respectively. 

Then  S  :  S'  :  :  A&  :  A7!?, 

(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 

\[S  :  v^  :  :  . iB  :  A' B'f  268 

or,  AB  :  A' B'  :  :  yfi  :  fl?. 


194 


GEOMETRY. BOOK  IV. 


On  Constructions. 

Proposition  XVI.     Problem. 

346.  To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares. 


B'h 

1  N 

Al X.._ 

s 

Rf 

R 

Let  R  and  R'  be  two  given  squares. 
It  is  required  to  construct  a  square  =  R+  R'. 
Construct  the  rt.  Z  A. 

Take  A  B  equal  to  a  side  of  Rf 

and  A  C  equal  to  a  side  of  R'. 

Draw  5(7. 

Then  B  C  will  be  a  side  of  the  square  required. 

For  130*  =  AB2  +  ~AC2,  §  331 

(the  square  on  the  hypotenuse  of  a  rt.  A  is  equivalent  to  the  sum  of  the 
squares  on  the  two  sides). 

Construct  the  square  S,  having  each   of   its  sides  equal 
to  BC. 

Substitute   for  BC2,   J~B*   and   jR?,  S,   R,   and  Rl  re- 
spectively ; 

then  8  =  R  +  R'. 

.*.  #  is  the  square  required. 

Q.  E.  F. 


CONSTRUCTIONS. 


195 


Proposition  XVII     Problem. 

347.   To  construct  a  square  equivalent  to  the  difference 
of  two  given  squares. 


\  *   I 


y<rx 


Let  R  be  the  smaller  square  and  R'  the  larger. 
It  is  required  to  construct  a  square  =  R!  —  R. 
Construct  the  rt.  Z.  A. 

Take  A  B  equal  to  a  side  of  R. 

From  B  as  a  centre,  with  a  radius  equal  to  a  side  of  Rf, 

describe  an  arc  cutting  the  line  A  X  at  C. 

Then  A  C  will  be  a  side  of  the  square  required. 

For  draw.SC. 

A-B*  +  A-C2  =  B7?,  §331 

{the  sum,  of  the  squares  on  the  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse). 

By  transposing,  AC1  =  BC2  —  AT?. 

Construct  the  square  S,  having  each  of  its  sides  equal  to  A  C. 

Substitute   for   AC*,  BC2,  and  IT?,    S,   R',  and   R  re- 
spectively ; 

then  S  =  R'  —  R. 


.*.  S  is  the  square  required. 


Q.  E.  F 


196  GEOMETRY.  —  BOOK   IV. 

Proposition  XVIII.     Problem. 

348.   To  construct  a  square  equivalent  to  the  sum  of  any 
number  of  given  squares. 


y\ 

s   \ 

F/ 

\ 
\ 

/  \ 

\ 

/     \ 

\ 

\        \ 
\ 
\ 
\    \ 

€t<^ 

\  \  \ 

-C:^ 

A^ 

j:-^« 

Let  m,  n,  o,  p,  r  be  sides  of  the  given  squares. 
It  is  required  to  construct  a  square  =  m?  +  n2  +  o2  +  p2  +  r3. 
Take  A  B  =  m. 

Draw  A  C  =  n  and  _L  to  A  B  at  A. 
Draw  B  C. 
Draw  C  E  =  o  and  i.  to  B  C  at  C,  and  draw  B  E. 
Draw  EE  =  ptmd  JL  to  BE  at  E,  and  draw  BE. 
Draw  EH  =  r  and  J_  to  B  E  at  F,  and  draw  B  H. 
The  square  constructed  on  B  H  is  the  square  required. 

For      BE2  =  FTP  +  £~F*, 

-  FTP  +  ^T2  +  ^g2, 

=  EH2  +  JTF2  +  E~G2  +  tf^2, 

=  #7Z2  +  EE2  +  .tftf2  +  cT  +  l^2,  §  331 

(the  sum  of  the  squares  on  two  sides  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse). 

Substitute  for  AB,  C  A,  EC,  E E,  and  F H,  m,  n}  o,  p, 
and  r  respectively; 

then  BH2  =  m2  +  n2  +  o2  +  p2  +  r*. 

Q.  E.  F. 


CONSTRUCTIONS. 


197 


Proposition  XIX.     Problem. 

849.   To  construct  a  polygon  similar  to  two  given  similaf 
polygons  and  equivalent  to  their  sum. 


Let  R  and  R'  be  two  similar  polygons,  and  AB  and 
A'  B'  two  homologous  sides. 

It  is  required  to  construct  a  similar  polygon  equivalent  to 
R+  R'. 

Construct  the  rt.  Z  P. 

Take  PH=  A' £',  and  P  0  =  A  B. 

Draw  0  H. 

Take  A"  B"  =  0  H. 

Upon  A"  B",  homologous  to  A  B,  construct  the  polygon  R" 
similar  to  R. 

Then  R"  is  the  polygon  required. 

For  R'  :  R  :  :  A1"^  :  ATff,  §  343 

{similar  polygons  ore  to  each  other  as  the  squares  on  their  homologous  sides). 


Also  R"  :  R'  :  :  A"  B"6  \  A'  B'\ 

In  the  first  proportion,  by  composition, 

R'  +  R  :  R'  ::  AHB*  +  Ftf  :  A1^'2, 
FH2  +  FO2  :  FE2, 
IH)2  :  PH\ 
:  AfrW1'1 


But 


R"  :  R' 


§343 
§  264 


A'B1 


:  ED2  :  PH2. 


\R"  :  R' 

.'.  R" 


R'+R 
R'  +  R. 


R'\ 


q.  e.  r. 


198 


GEOMETRY. 


BOOK   IV. 


Proposition  XX.     Problem. 

350.   To  construct  a  polygon  similar  to  two  given 
polygons  and  equivalent  to  their  difference. 


/     •  I X.... 

A'  B'  A  B  A'f         Bii       P  0 

Let  R  and  R'  be  two  similar  polygons,  and  AB  and 

A' B'   two  homologous  sides. 

It  is  required  to  construct  a   similar  polygon  which  shall 
be  equivalent  to  R'  —  R. 

Construct  the  rt.  Z  P, 

and  take  PO  =  AB. 

From  0  as  a  centre,  with  a  radius  equal  to  A'  B'> 

describe  an  arc  cutting  P  X  at  H. 

Draw  0  H. 

Take  A"  B"  =  P  H. 

On  A"  B",  homologous  to  A  B,  construct  the  polygon  R" 
similar  to  R. 

Then  R"  is  the  polygon  required. 

For  R'  :  R  :  :  A7^  :  AB*,  §  343 

(similar  polygons  are  to  each  other  as  the  squares  on  their  homologous  sides). 

§343 
§  265 


Also  R"  :  R  :  :  A"  B"'  :  A  B\ 

In  the  first  proportion,  by  division, 


R  :  R 


But 


R"  :  R 


A1^  -  AB2  :  ITS', 

on2  -  op*  :  oy*, 

PH2  :  UP*. 

2  yJT&t 


:  A"  B 

:  PH2  :  U~P. 
.R".R.:R'-R:R', 
.'.  R"  =  R1  —  R. 


Q.  E.  F. 


CONSTRUCTIONS. 


199 


polygon. 


Proposition  XXI.     Problem. 
351.    To   construct  a   triangle   equivalent   to    a  given 


I  A  E  F 

Let  ABC DH E  be  the  given  polygon. 

It  is  required  to  construct  a  triangle  equivalent  to  the  given 
polygon. 

From  D  draw  D  E,  and  from  H  draw  HF  II  to  D  E. 

Produce   A  E  to  meet  HF  at  F,  and  draw  D  F. 

The  polygon  A  BG D F  has  one  side  less  than  the  polygon 
ABC D H E,  but  the  two  are  equivalent. 

For  the  part  A  B  CD  E  is  common, 

and  the  A  D  EF=  A  D  EH,  fox  the  base  D  E is  common, 
and  their  vertices  F and  H  are  in  the  line  FH II  to  the  base,     §  325 
(/&  having  the  same  base  and  equal  altitudes  are  equivalent). 

Again,  draw  C  F,  and  draw  D  K  II  to  0  F  to  meet  A  F 
produced  at  K. 

Draw  GK. 

The  polygon  ABG K  has  one  side  less  than  the  polygon 
ABG D  F,  but  the  two  are  equivalent. 

For  the  part  A  B  GF  is  common, 

and  the  A  G  FK  =  A  GFD,  for  the  base  GF  is  common, 
and  their  vertices  K  and  D  are  in  the  line  KD  II  to  the  base.      §  325 

In  like  manner  we  may  continue  to  reduce  the  number  of 
sides  of  the  polygon  until  we  obtain  the  A  G I K. 

Q.  E.  F. 


200  GEOMETRY.  —  BOOK   IV. 


Proposition  XXII.     Problem. 

352.  To  construct  a  square  which  shall  have  a  given 
ratio  to  a  given  square. 

S 


.--"""TV-.. 
\ 


m fa. 


\    \ 


^o 


n 
Let  R  be  the  given  square,  and  -  the  given  ratio. 

m 

It  is  required  to  construct  a  square  which  shall  be  to  R  as 
n  is  to  m. 

On  a  straight  line  take  AB  =  m,  and  BC  =  n. 

On  A  G  as  a  diameter,  describe  a  semicircle. 

At  B  erect  the  J_  B S,  and  draw  SA  and  SG. 

Then  the  A  A  S  G  is  a  rt.  A  with  the  rt.  Z  at  S,     §  204 
(being  inscribed  in  a  semicircle. ) 

On  S  A,  oi  S  A  produced,  take  SE  equal  to  a  side  of  R. 

Draw  EF  li  io  AG. 

Then  S  F  is  a  side  of  the  square  required. 

For  S-'^'i  §289 

(£Ae  squares  on  the  sides  of  a  rt.  A  have  the  same  ratio  as  the  segments  of  the 
hypotenuse  made  by  the  JL  let  fall  from  the  vertex  of  the  rt.  Z). 

Also  M  » :**i  §  275 

SG        SF'  * 

(a  straight  line  dravm  through  two  sides  of  a  A,  parallel  to  the  third  side, 
divides  those  sides  proportionally). 

Square  the  last  equality ; 

then  12.5* 

5T2       ST* 

ft   J2  <V  W2 

Substitute,  in  the  first  equality,  for  its  equal  j 

then  ^  =  ^=-, 

S~F2        BG        n 

that  is,  the  square  having  a  side  equal  to  SF  will  have  the 
same  ratio  to  the  square  R,  as  n  has  to  m. 

Q.  E.  F. 


CONSTRUCTIONS.  201 


Proposition  XXIII.     Problem. 

853.  To  construct  a  polygon  similar  to  a  given  polygon 
and  having  a  given  ratio  to  it. 


m v  / 


n 


At  & 


n 
Let  R  be  the  given  polygon  and  -  the  given  ratio. 

m 

It  is  required  to  construct  a  polygon  similar  to  R,  which 
shall  be  to  R  as  n  is  to  m. 

Find  a  line,  A'  B',  such  that  the  square  constructed  upon  it 
shall  be  to  the  square  constructed  upon  A  B  as  n  is  to  m.     §  352 

Upon  A'  B'  as  a  side  homologous  to  A  B,  construct  the 
polygon  S  similar  to  R. 

Then  S  is  the  polygon  required. 
S!         A7!?'2 

(similar  polygons  are  to  each  other  as  the  squares  on  their  homologous  sides). 

But  ^  _  -  J  Cons. 

ATtf        m 

.*.—  =  _,     or,     S  :  R  '.  :  n  :  m. 
R        m 

Q.  E.  F 


202  GEOMETRY. BOOK   IV. 


Proposition  XXIV.     Problem. 

354.  To  construct  a  square  equivalent  to  a  given  paral- 
lelogram. 

P 

B  C   r ]  ^"T N 

/    \ 

"  i  /      \ 

A  b  D         J    Ml N g-3C 

Let  ABCD  be  a  parallelogram,  b  >.bs  base,  and  a  its 
altitude. 

It  is  required  to  construct  a  square  =  EJ  A  B  C  D. 

Upon  the  line  MX  take  M N  =  a,  and  N  0  =  b. 

Upon  if  0  as  a  diameter,  describe  a  semicircle. 

At  N  eiect  NP±  to  MO. 

Then  the  square  R,  constructed  upon  a  line  equal  to  N P, 
is  equivalent  to  the  O  A  B  G  D. 

For  MN  :  NP  :  :  NP  :  NO,  §  307 

(a  A.  let  fall  from  any  point  of  a  circumference  to  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

.'.  NP*  =  MN  X  NO  =  aXb,  §259 

(the  product  of  the  means  is  equal  to  the  product  of  the  extremes). 

Q.  E.  F. 

355.  Corollary  1.  A  square  may  be  constructed  equiva- 
lent to  a  triangle,  by  taking  for  its  side  a  mean  proportional 
between  the  base  and  one-half  the  altitude  of  the  triangle. 

356.  Cor.  2.  A  square  may  be  constructed  equivalent  to 
any  polygon,  by  first  reducing  the  polygon  to  an  equivalent  tri- 
angle, and  then  constructing  a  square  equivalent  to  the  triangle. 


CONSTRUCTIONS.  203 


Proposition  XXY.     Problem. 

357.  To  construct  a  parallelogram  equivalent  to  a  given 
square,  and  having  the  sum  of  its  base  and  altitude  equal  to 
a  given  line. 

Ss" 


It 


A      ^ 

if  LX J  n 

Let  R  be  the  given  square,  and  let  the  sum  of  the 
base  and  altitude  of  the  required  parallelogram 
be  equal  to  the  given  line  M N. 

It  is  required  to  construct  a  O  =  R,  and  having  the  sum 
of  its  base  and  altitude  =  M  N. 

Upon  M  N  as  a  diameter,  describe  a  semicircle. 
At  M  erect  a  J_  MP,  equal  to  a  side  of  the  given  square  JR. 
Draw  P  Q  II  to  M N,  cutting  the  circumference  at  S. 
Draw  SO±  to  M N. 
Any  O  having  G  M  for  its  altitude  and  G  N  for  its  base, 
is  equivalent  to  R. 

For  tftfisll  toPJf,  §65 

{two  straight  lines  _L  to  the  same  straight  line  are  II ). 

.\SC  =  PM,  §135 

(lis  comprehended  between  lis  arc  equal). 

.'.  Slf  =  PM2  =  R. 

But  MG  :  SC  :  :  SC  :  C N,  §307 

(a  _L  let  fall  from  any  point  in  a  circumference  to  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

Then  SGl  =  MCXGN,  §259 

(the  product  of  the  means  is  equal  to  the  product  of  the  extremes). 


204 


GEOMETRY. BOOK  IV, 


Proposition  XXVI.     Problem. 

859.   To  construct  a  parallelogram  equivalent  to  a  given 
square,  and  having  the  difference  of  its  base  and  altitude 
equal  to  a  given  line. 
S 


\C 


M 


r 


r 


-/ 


R> 


\ 


>N 


/ 


B 

Let  R  be  the  given  square,  and  let  the  difference  of 
the  base  and  altitude  of  the  required  parallelo- 
gram be  equal  to  the  given  line  M N. 

It  is  required  to  construct  a  E3  =  R,  with  the  difference 
of  the  base  and  altitude  =  M  N. 

Upon  the  given  line  M  N  as  a  diameter,  describe  a  circle. 

From  M  draw  MS,  tangent  to  the  O,  and  equal  to  a  side 
of  the  given  square  R. 

Through  the  centre  of  the  O,  draw  #2?  intersecting  the 
circumference  at  G  and  B. 


Then  any  O,  as  R',  having  SB  for  its  base  and  SC  for 
its  altitude,  is  equivalent  to  R. 

For  SB  :  SM  :  :  SM  :  S  C,  §  292 

(if  from  a  point  without  a  O,  a  secant  and  a  tangent  be  drawn,  the  tangent  is 
a  mean  proportional  between  the  whole  secant  and  the  part  without  the  O). 

Then  8~M2  =  SBXSC;  §259 

and  the    difference  between  SB  and  SC  is  the  diameter 
of  the  O,  that  is,  M N. 

Q.  E.  F. 


CONSTRUCTIONS.  205 


Proposition  XXVII.     Problem. 


860.   Given  x  =  fa,  to  construct  x. 


E 


x, 


D    B 


Let  m  represent  the  unit  of  length. 

It  is  required  to  find  a  line  which  shall  represent  the  square 
root  of  2. 

On  the  indefinite  line  A  B,  take  A  C  =  m,  and  CD  =  2  m. 

On  A  D  as  a  diameter  describe  a  semi-circumference. 

At  C  erect  a  JL  to  A  B,  intersecting  the  circumference  at  E. 

Then  C  E  is  the  line  required. 

For  AC  :  CE  :  :  CE  :  CD,  §  307 

{the  ±  let  fall  from  any  point  in  the  circumference  to  the  diameter,  is  a  mean 
proportional  between  the  segments  of  the  diameter)  ; 

.-.C~E2  =  ACX  CD,  §259 


.\CE=slACX  CD, 
=  vfTx~2  =  fa . 


Q.  E.  F. 


Ex.   1.    Given  x  =  y/5,  y  =  y/7,  z  —  2  fa> ;  to  construct  x,  y, 
and  z. 

2.  Given  2   :  x  :  :  x  :  3 ;  to  construct  #. 

3.  Construct  a  square  equivalent  to  a  given  hexagon. 


206  GEOMETRY. BOOK   IV. 


Proposition  XXVIII.     Problem. 

361.  To  construct  a  polygon  similar  to  a  given  polygon 
P,  and  equivalent  to  a  given  polygon  Q. 

A 

\        P'        I 

A'  B' 


m      A  "      "b 

Let  P  and  Q  be  two  given  polygons,  and  A  B  a  side 
of  polygon  P. 

It  is  required  to  construct  a  polygon  similar  to  P  and  equiva- 
lent to  Q. 

Find  a  square  equivalent  to  Pf  §  356 

and  let  m  be  equal  to  one  of  its  sides. 

Find  a  square  equivalent  to  Q,  §  356 

and  let  n  be  equal  to  one  of  its  sides. 

Find  a  fourth  proportional  to  m,  n,  and  A  B.  §  304 

Let  this  fourth  proportional  be  A'  B'. 

Upon  A1  B',  homologous  to  A  B,  construct  the  polygon  F 
similar  to  the  given  polygon  P. 

Then  P'  is  the  polygon  required. 


CONSTRUCTIONS.  207 


For  ™  =  —  •  Cons. 

n         A'B' 


Squaring, 

m>         AB' 

But 

P  =  m?, 

and 

Q  =  n*; 

.    P 

"  Q 

m2     r& 

P          A& 

Cons. 
Cons. 


But  T,  =  ^>  §343 

P>  JTff* 

(similar  polygons  are  to  each  other  as  the  squares  on  their  homologous  sides)  ; 

.-.-  =  — ;  Ax.  1 

Q       P' 

.'.  P'  is  equivalent  to  Q,  and  is  similar  to  P  by  construction. 

Q.  E.  F. 


Ex.  1.  Construct  a  square  equivalent  to  the  sum  of  three 
given  squares  whose  sides  are  respectively  2,  3,  and  5. 

2.  Construct  a  square  equivalent  to  the  difference  of  two 
given  squares  whose  sides  are  respectively  7  and  3. 

3.  Construct  a  square  equivalent  to  the  sum  of  a  given  tri- 
angle and  a  given  parallelogram. 

4.  Construct  a  rectangle  having  the  difference  of  its  base  and 
altitude  equal  to  a  given  line,  and  its  area  equivalent  to  the  sum 
of  a  given  triangle  and  a  given  pentagon. 

5.  Given  a  hexagon ;  to  construct  a  similar  hexagon  whose 
area  shall  be  to  that  of  the  given  hexagon  as  3  to  2. 

6.  Construct   a   pentagon  similar  to  a  given  pentagon  and 

equivalent  to  a  given  trapezoid. 


208  GEOMETEY. BOOK   IV. 


Proposition  XXIX.     Problem. 

862.  To  construct  a  polygon  similar  to  a  given  polygon, 
and  having  two  and  a  half  times  its  area. 

Y 


A/I 

B  M        C  0  N 

Let  P  be  the  given  polygon. 

It   is   required   to   construct   a  polygon   similar   to  P,  and 
equivalent  to  2  J  P. 

Let  A  B  be  a  side  of  the  given  polygon  P. 
Then  ^T  :  s/T%  :  :  A  B  :  x, 

or  \J2   :  \Jd   :  :  A  B  :  x,  §  345 

{the  homologous  sides  of  similar  polygons  are  to  each  other  as  the  square  roots 
of  their  areas). 

Take  any  convenient  unit  of  length,  as  MC,  and  apply  it 
six  times  to  the  indefinite  line  M  N. 

On  M  0  (=  3  M  C)  describe  a  semi-circumference ; 

and  on  M  N  (=  6  M  C)  describe  a  semi-circumference. 

At  C  erect  a  _L  to  M  N,  intersecting  the  semi-circumfer- 
ences at  D  and  H. 

Then  CD  is  the  sj%  and  0 H  is  the  \/E.  §  360 

Draw  C  7,  making  any  convenient  Z  with  C  H. 
On  CY  take  C  E  =  A  B. 
From  D  draw  D  E, 
and  from  H  draw  H  Y  \\  to  D  E. 


CONSTRUCTIONS. 


209 


Then  C  Y  will  equal  x,  and  be  a  side  of  the  polygon  re- 
quired, homologous  to  A  B. 

For  CD  :  CH  :  :  CE  :  C  Y,  §275 

(a  line  drawn  through  two  sides  of  a  A,  II  to  the  third  side,  divides  the  two 
sides  proportionally). 

Substitute  their  equivalents  for  CD,  C H,  and  C E ; 

then  \J2   :  V^  :  :  A  B  :  C  Y. 

On  C  Y,  homologous  to  A  B,  construct  a  polygon  similar 
to  the  given  polygon  P ; 

and  this  is  the  polygon  required. 

Q.  E.  F. 


Ex.  1.  The  perpendicular  distance  between  two  parallels  is 
30,  and  a  line  is  drawn  across  them  at  an  angle  of  45° ;  what  is 
its  length  between  the  parallels  1 

2.  Given  an  equilateral  triangle  each  of  whose  sides  is  20 ; 
find  the  altitude  of  the  triangle,  and  its  area. 

3.  Given  the  angle  A  of  a  triangle  equal  to  |  of  a  right 
angle,  the  angle  B  equal  to  J  of  a  right  angle,  and  the  side  a, 
opposite  the  angle  A,  equal  to  10 ;  construct  the  triangle. 

4.  The  two  segments  of  a  chord  intersected  by  another  chord 
are  6  and  5,  and  one  segment  of  the  other  chord  is  3 ;  what 
is  the  other  segment  of  the  latter  chord  % 

5.  If  a  circle  be  inscribed  in  a  right  triangle  :  show  that 
the  difference  between  the  sum  of  the  two  sides  containing  the 
right  angle  and  the  hypotenuse  is  equal  to  the  diameter  of  the 
circle. 

6.  Construct  a  parallelogram  the  area  and  perimeter  of  which 
shall  be  respectively  equal  to  the  area  and  perimeter  of  a  given 
triangle. 

7.  Given  the  difference  between  the  diagonal  and  side  of  a 
square;  construct  the  square. 


BOOK  V. 

REGULAR  POLYGONS  AND   CIRCLES. 

363.  Def.    A   Regular   Polygon    is    a    polygon    which    is 
equilateral  and  equiangular. 

Proposition  I.     Theorem. 

364.  Every  equilateral  polygon  inscribed  in  a  circle  is  a 
regular  polygon. 


Let  ABC,  etc.,    be  an    equilateral  polygon  inscribed 
in  a  circle. 

We  are  to  prove  the  polygon  ABC,  etc.,  regular. 

The  arcs  A  B,  B  C,  C  D,  etc.,  are  equal,  §  182 

(in  the  same  O,  equal  chords  subtend  equal  arcs). 

.'.  arcs  ABC,  BCD,  etc.,  are  equal,  Ax.  6 

.*.  the  A  A,  B,  C,  etc.,  are  equal, 
(being  inscribed  in  equal  segments). 

.'.  the  polygon   ABC,  etc.,   is  a   regular   polygon,   being 
equilateral  and  equiangular. 

Q.  E.  D. 


REGULAR   POLYGONS    AND    CIRCLES.  211 

Proposition  II.     Theorem. 

365.  I.    A  circle  may  be  circumscribed  about  a  regular 
polygon. 

II.  A  circle  may  be  inscribed  in  a  regular  polygon. 


Let  A  BCD,  etc.,  be  a  regular  polygon. 
We  are  to  prove  that  a  O  mag  be  circumscribed  about  this 
regular  polygon,  and  also  a  O  mag  be  inscribed  in  this  regular 
polygon. 

Case  I.  —  Describe  a  circumference  passing  through  A,  B,  and  C. 

From  the  centre  0,  draw  0  A,  0  D, 

and  draw  0  s  _L  to  chord  B  C. 

On  0  s  as  an  axis  revolve  the  quadrilateral  0  ABs, 

until  it  comes  into  the  plane  of  0 sC D. 

The  line  s  B  will  fall  upon  s  C, 
(for  ZOsB  =  ZOsC,  both  being  rt.  A ). 

The  point  B  will  fall  upon  C,  §  183 

(since  s  B  —  s  C). 

The  line  BA  will  fall  upon  CD,  §  363 

(since  /.  B  =  Z  C,  being  A  of  a  regular  polygon). 

The  point  A  will  fall  upon  D,  §  363 

(since  B  A  =  C  D,  being  sides  of  a  regular  polygon). 

.'.  the  line  0  A  will  coincide  with  line  0  D, 
(their  extremities  being  the  same  points). 

.'.  the  circumference  will  pass  through  D. 
In  like  manner  we  may  prove  that  the  circumference,  pass- 
ing through  vertices  B,  C,  and  D  will  also  pass  through  the 
vertex  E,  and  thus  through  all  the  vertices  of  the  polygon  in 
succession. 

Case  II. — The  sides  of  the  regular  polygon,  being  equal  chords  of 

the  circumscribed  O,  are  equally  distant  from  the  centre,     §  1 85 

.'.a  circle  described  with  the  centre  0  and  a  radius  Os 

will  touch  all  the  sides,  and  be  inscribed  in  the  polygon.     §  174 


212  GEOMETRY. BOOK  V. 

366.  Def.    The  Centre  of  a  regular  polygon  is  the  common 
centre  0  of  the  circumscribed  and  inscribed  circles. 

367.  Def.    The  Radius  of  a  regular  polygon  is  the  radius 
0  A  of  the  circumscribed  circle. 

368.  Def.    The  Apothem  of  a  regular  polygon  is  the  radius 
0  s  of  the  inscribed  circle. 

369.  Def.    The  Angle  at  the  centre  is  the  angle  included 
by  the  radii  drawn  to  the  extremities  of  any  side. 


Proposition  III.     Theorem. 

370.  Each  angle  at  the  centre  of  a  regular  polygon  is 
equal  to  four  right  angles  divided  by  the  number  of  sides 
of  the  polygon. 


B 
Let 'ABC,  etc.,  be  a  regular  polygon  of  n  sides. 

_  4  rt.  A 
We  are  to  prove      Z.  A  0  B  —  — —       • 

Circumscribe  a  O  about  the  polygon. 

The  AAOB,BOC,  etc.,  are  equal,  §  180 

(in  the  same  O  equal  arcs  subtend  equal  A  at  tJie  centre). 

.'.  the  Z  A  0  B  =  4  rt.  A  divided  by  the  number  of  A  about  0. 

But  the  number  of  A  about  0  =  n,  the  number  of  sides 
of  the  polygon. 

4  rt.  A 

*******  nr» 

Q.  E.  D. 

371.  Corollary.    The   radius   drawn  to  any  vertex  of  a 
regular  polygon  bisects  the  angle  at  that  vertex. 


REGULAR   POLYGONS    AND    CIRCLES.  213 


Proposition  IV.     Theorem. 

372.   Two  regular  polygons  of  the  same  number  of  sides 
are  similar. 


Let  Q  and  Q'  be  two  regular  polygons,  each  having 
n  sides. 

We  are  to  prove       Q  and  Q1  similar  polygons. 

The   sum  of  the  interior  A  of  each  polygon  is  equal  to 
2  rt.  A  (n  -  2),  §  157 

(tlie  sum  of  tlie  interior  A  of  a  polygon  is  equal  to  2  rt.  A  taken  as  many 
times  less  2  as  the  polygon  has  sides). 

Each  A  of  the  polygon  Q  = — '  >         §  158 

(for  the  A  of  a  regular  polygon  are  all  equal,  and  hence  each  Z  is  equal 
to  the  sum  of  the  A  divided  by  their  number). 

Also,  each  A  of  Q'  =  2  rt  A(n~2) .  §  153 

n 

.'.  the  two  polygons  Q  and  Q'  are  mutually  equiangular. 

Moreover,  =  ^  5  363 

(the  sides  of  a  regular  polygon  are  all  equal) ; 

and  ^JL  =  1,  §  3G3 

B'C 

...£*«£*,  Ax.l 

B  G        B'C 

.'.  the  two  polygons  have  their  homologous  sides  proportional ; 

.'.  the  two  polygons  are  similar.  §  278 

Q.  E.  D. 


214 


GEOMETRY. 


BOOK    V. 


Proposition  V.     Theorem. 

373.  The  7wmologous  sides  of  similar  regular  polygons 
have  the  same  ratio  as  the  radii  of  their  circumscribed  cir- 
cles, and,  also  as  the  radii  of  their  inscribed  circles. 


Let  0  and  0'  be  the  centres  of  the  two  similar  regu- 
lar polygons  ABC,  etc.,  and  A'B'C,  etc. 

From  0  and   0'  draw  0  E,   0  D,   O'E',   0' D',   also   the 
Js  0  m  and   0'  m'. 

0  E  and  0'  E'  are  radii  of  the  circumscribed  (D,         §  367 

and  Om  and  O'm'  are  radii  of  the  inscribed  (D.         §  368 

ED  OE  Om 

=  O'E  " 


We  are  to  prove 


ED'        O'E'        O'm' 

In  the  AOED  and  0' E> D' 

the  A  0  E  D,  0  D  E,  0'  E  D'  and  0'  D'  E'  are  equal,     §  371 
{being  halves  of  the  equal  A  F  E  D,  E  D  C,  F'  E>  Df  and  E>  D'  O)  ; 

.'.  the  A  0  ED  and  0'  E  D'  are  similar,  §  280 

(if  two  A  have  two  A  of  the  one  equal  respectively  to  two  A  of  the  other,  they 
are  similar). 


ED 


OE 


Also, 


E'D'        O'E' 
(the  homologous  sides  of  similar  A  are  proportional). 

ED  Om 


E'D' 


O'm' 


§278 


§297 


(the  homologous  altitudes  of  similar  A  have  the  same  ratio  as  their  homolo- 

qous  bases). 

Q.  E.  D. 


REGULAR    POLYGONS    AND    CIRCLES. 


215 


Proposition  VI.     Theorem. 

374.  The  perimeters  of  similar  regular  polygons  have 
the  same  ratio  as  the  radii  of  their  circumscribed  circles,  and. 
also  as  the  radii  of  their  inscribed  circles. 

A>^- ~-\B' 


Let  P  and  P'  represent  the  perimeters  of  the  two 
similar  regular  polygons  ABC,  etc.,  and  A'B'C,  etc. 
From   centres  0,  0'  draw  0  E,  0'  E',  and  J§  0  m  and  0'  m'. 

Om 


P  OP 

We  are  to  prove      —  =  

1  P>        0>  E' 


ED 


O'm' 


§  295 
F        E'  D'  * 

(the  perimeters  of  similar  polygons  have  the  same  ratio  as  any  two  homolo- 


gous 


Moreover, 


OE         ED 
0> E'  ~~  EH)'9 


(the  homologous  sides  of  similar  regular  polygons  have  the 
radii  of  their  circumscribed  (D). 


Also 


0 


O'm' 


ED 
~ErD}i 


§373 
ratio  as  the 

§373 


(the  homologous  sides  of  similar  regulmr  polygons  have  the  same  ratio  as 
the  radii  of  their  inscribed  (D). 


OE 
O'E' 


Om 

Wm' 


Q.  E.  D. 


216 


GEOMETRY. BOOK    V. 


Proposition  VII.     Theorem. 

375.  The  circumferences  of  circles  have  the  same  ratio 
as  their  radii. 


Let  C  and  C  be    the    circumferences,  R  and  R'  the 
radii  of  the  two  circles  Q  and  Q'. 

We  are  to  prove      G  :  C  :  :  R  :  R'. 

Inscribe  in  the  (D  two  regular  polygons  of  the  same  number 
of  sides. 

Conceive  the  number  of  the  sides  of  these  similar  regular 
polygons  to  be  indefinitely  increased,  the  polygons  continuing  to 
be  inscribed,  and  to  have  the  same  number  of  sides. 

Then  the  perimeters  will  continue  to  have  the  same  ratio  as 
the  radii  of  their  circumscribed  circles,  §  374 

(the  perimeters  of  similar  regular  polygons  have  the  same  ratio  as  the  radii 
of  their  circumscribed  (D), 

and  will  approach  indefinitely  to  the  circumferences  as  their 
limits. 

.'.  the  circumferences  will  have  the  same  ratio  as  the  radii 
of  their  circles,  §  1 99 

.'.C  :  C0::  R  :  R'. 

Q.  E.  O 


REGULAR   POLYGONS    AND    CIRCLES.  217 

376.  Corollary.  By  multiplying  by  2,  both  terms  of  the 
ratio  R  :  R',  we  have 

0 ;  C  : :  2  R  i  2  #  j 

that  is,  the  circumferences  of  circles  are  to  each  other  as 
their  diameters. 

Since  C  :  C  :  :  2  R  :  2  R', 

G  :  2  7?  :  :  C  :  2  R',  §  262 

or,  =  — —  . 

2  R        2R' 

That  is,  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  a  constant  quantity. 

This  constant  quantity  is  denoted  by  the  Greek  letter  ir. 

377.  Scholium.  The  ratio  tt  is  incommensurable,  and  there- 
fore can  be  expressed  only  approximately  in  figures.  The  let- 
ter 77,  however,  is  used  to  represent  its  exact  value. 


Ex.  1.  Show  that  two  triangles  which  have  an  angle  of  the 
one  equal  to  the  supplement  of  the  angle  of  the  other  are  to  each 
other  as  the  products  of  the  sides  including  the  supplementary 
angles. 

2.  Show,  geometrically,  that  the  square  described  upon  the 
sum  of  two  straight  lines  is  equivalent  to  the  sum  of  the  squares 
described  upon  the  two  lines  plus  twice  their  rectangle. 

3.  Show,  geometrically,  that  the  square  described  upon  the 
difference  of  two  straight  lines  is  equivalent  to  the  sum  of  the 
squares  described  upon  the  two  lines  minus  twice  their  rectangle. 

4.  Show,  geometrically,  that  the  rectangle  of  the  sum  and 
difference  of  two  straight  lines  is  equivalent  to  the  difference 
of  the  squares  on  those  lines. 


218  GEOMETRY. BOOK    V. 


Proposition  VIII.     Theorem. 

378.  If  the  number  of  sides  of  a  regular  inscribed  poly- 
gon be  increased  indefinitely,  the  apothem,  will  be  an  increas- 
ing variable  whose  limit  is  the  radius  of  the  circle. 


In  the  right  triangle  OCA,  let  0  A  he  denoted  by  R, 
OC  byr,  and  A  G  by  b. 

We  are  to  prove  Urn.  (r)  =  R. 

r<R,  §52 

(a  _L  is  the  shortest  distance  from  a  point  to  a  straight  line). 

And  R-r<b,  §97 

{one  side  of  a  A  is  greater  than  the  difference  of  the  other  two  sides). 

By  increasing  the  number  of  sides  of  the  polygon  indefi- 
nitely, A  B,  that  is,  2  b,  can  be  made  less  than  any  assigned 
quantity. 

.'.  b,  the  half  of  2  b,  can  be  made  less  than  any  assigned 
quantity. 

.'.  R  —  r,  which  is  less  than  b,  can  be  made  less  than  any 
assigned  quantity. 

.'.Urn.  (R  —  r)  =  0. 

.'.R-lim.  (r)  =  0.  §  199 

.*.  Urn.  (r)  =  R. 

Ql.  e.  d. 


REGULAR   POLYGONS    AND    CIRCLES. 


219 


Proposition  IX.     Theorem. 

379.   The  area  of  a  regular  polygon  is  equal  to  one-half 
the  product  of  its  apothem  by  its  perimeter. 

B 


Let  P  represent  the  perimeter  and  R    the  apothem 
of  the  regular  polygon  ABC,  etc. 

We  are  to  prove      the  area  of  A  BG,  etc.,  =  J  R  X  P. 

Draw  OA,  OB,  OC,  etc. 

The  polygon  is  divided  into  as  many  A  as  it  has  sides. 

The  apothem  is  the  common  altitude  of  these  A, 

and   the  area  of  each   A  is  equal  to  J  R  multiplied  by 
the  base.  §  324 

.*.  the  area  of  all  the  A  is  equal  to  J  R  multiplied  by  the 
sum  of  all  the  bases. 

But  the  sum  of  the  areas  of  all  the  A  is  equal  to  the  area 
of  the  polygon, 

and  the  sum  of  all  the  bases  of  the  A  is  equal  to  the 
perimeter  of  the  polygon. 

.".  the  area  of  the  polygon  =  J  R  X  P. 

Q.  E.  D. 


220  GEOMETRY. BOOK    V. 


Proposition  X.     Theorem. 

380.    The  area   of  a   circle   is   equal  to   one-half  the 
product  of  its  radius  by  its  circumference. 


Let  R  represent  the  radius,  and  0  the  circumference 
of  a  circle. 

We  are  to  prove      the  area  of  the  circle  =  J  R  X  C. 

Inscribe  any  regular  polygon,  and  denote  its  perimeter 
by  P,  and  its  apothem  by  r. 

Then  the  area  of  this  polygon  =  J  r  X  P,  §379 

(the  area  of  a  regular  polygon  is  equal  to  one-hulf  the  product  of  its  apothem 
by  the  perimeter). 

Conceive  the  number  of  sides  of  this  polygon  to  be  indefi- 
nitely increased,  the  polygon  still  continuing  to  be  regular  and 
inscribed. 

Then  the  perimeter  of  the  polygon  approaches  the  circum- 
ference of  the  circle  as  its  limit, 

the  apothem,  the  radius  as  its  limit,  §  378 

and  the  area  of  the  polygon  approaches  the  O  as  its  limit. 

But  the  area  of  the  polygon  continues  to  be  equal  to  one- 
half  the  product  of  the  apothem  by  the  perimeter,  however 
great  the  number  of  sides  of  the  polygon. 

.'.  the  area  of  the  O  =  J  R  X  0.  §  199 

Q.  E.  D. 

I 


REGULAR   POLYGONS    AND    CIRCLES.  221 

c 

381.  Corollary  1.    Since  —  as  »«  $  376 

2  R  * 

.-.  C=2ttE. 
In  the  equality,  the  area  of  the  O  =  \  R  X  C, 
substitute  2  ir  R  for  C ; 
then  the  area  of  the  O  =  \  R  X  2  tt  i?, 

That  is,  ?Ae  area  o/  a  O  =  tt  tames  ?Ae  sgware  ora  tfc  radius. 

382.  Cor.  2.  7%e  area  o/  a  sector  eawafe  |  <Ae  product  of 
its  radius  by  its  arc ;  for  the  sector  is  such  part  of  the  circle  as 
its  arc  is  of  the  circumference. 

383.  Def.  In  different  circles  similar  arcs,  similar  sectors, 
and  similar  segments,  are  such  as  correspond  to  equal  angles  at 
the  centre. 

Proposition  XI.     Theorem. 

384.  Two  circles  are  to  each  other  as  the  squares  on 
their  radii. 


Let  R  and  R'  be  the  radii  of  the  two  circles  Q  and  Q'. 

We  are  to  prove      —    =  —  . 
*  Q'       R'2 

Now  Q  =  ttR2,  §381 

(the  area  of  aO  =  ir  times  the  square  on  its  radius), 

and  Q,  =  7rR/2.  §381 

Q        x  /.'■-'     _  m^ 

Q.  E.  D. 


Then  «  * 


385.  Corollary.  Similar  arcs,  being  like  parts  of  their  re- 
spective circumferences,  are  to  each  other  as  their  radii ;  similar 
sectors,  being  like  parts  of  their  respective  circles,  are  to  each 
other  as  the  squares  on  their  radii. 


222 


GEOMETRY. BOOK  V. 


Proposition  XII.     Theorem. 

386.  Similar  segments  are  to  each  other  as  the  squares 

on  their  radii.  C 

O 


P'  p 

Let  A  C  and  A' C  be  the  radii  of  the  two  similar  seg- 
ments ABP  and  A' B' P'. 

w  *  AB?         AC1 

We  are  to  prove 


A'B'P'        j[rQfl 

The  sectors  A  CB  and  A'  C B>  are  similar,  §  383 

{having  the  A  at  the  centre,  C  and  O,  equal). 

In  the  AACB  and  A'  C  B' 

£C  =  /.C,  §  383 

{being  corresponding  A  of  similar  sectors). 

AC=CB,  §  163 

A'C'  =  C'B';  §163 

.-.  the  A  A  C  B  and  A'  C  B'  are  similar,  §  284 

{having  an  4-  of  the  one  equal  to  an  Z.  of  the  other,  and  the  including  sides 
'proportional). 

Now  sector  ACB    _   AC2  §  385 

sector  A'C'B'        jrrjt 
{similar  sectors  are  to  each  other  as  the  squares  on  their  radii)  ; 

and  AACB   =A°L,  §342 


A  A'C'B' 


A'C 


{similar  A  are  to  each  other  as  the  squares  on  their  homologous  sides). 

-rr  sector  ACB-  A  ACB  AC2 

Hence  =  -  , 

sector  A'  C  B'  -  A  A'  C  B'        j7!?2 

or,  segment  A  BP   =   IV2  .  j  271 

segment  A'  B'  P'        AJU'2 
{if  two  quantities  be  increased  or  diminished  by  like  parts  of  each,  the  results 
will  be  in  the  same  ratio  as  the  quantities  themselves). 

Q.  E.  D. 


EXERCISES.  223 


Exercises. 

1.  Show  that  an  equilateral  polygon  circumscribed  about  a 
circle  is  regular  if  the  number  of  its  sides  be  odd. 

2.  Show  that  an  equiangular  polygon  inscribed  in  a  circle  is 
regular  if  the  number  of  its  sides  be  odd. 

3.  Show  that  any  equiangular  polygon  circumscribed  about  a 
circle  is  regular. 

4.  Show  that  the  side  of  a  circumscribed  equilateral  triangle 
is  double  the  side  of  an  inscribed  equilateral  triangle. 

5.  Show  that  the  area  of  a  regular  inscribed  hexagon  is 
three-fourths  of  that  of  the  regular  circumscribed  hexagon. 

6.  Show  that  the  area  of  a  regular  inscribed  hexagon  is  a 
mean  proportional  between  the  areas  of  the  inscribed  and  cir- 
cumscribed equilateral  triangles. 

7.  Show  that  the  area  of  a  regular  inscribed  octagon  is  equal 
to  that  of  a  rectangle  whose  adjacent  sides  are  equal  to  the 
sides  of  the  inscribed  and  circumscribed  squares. 

8.  Show  that  the  area  of  a  regular  inscribed  dodecagon  is 
equal  to  three  times  the  square  on  the  radius. 

9.  Given  the  diameter  of  a  circle  50 ;  find  the  area  of  the 
circle.     Also,  find  the  area  of  a  sector  of  80°  of  this  circle. 

10.  Three  equal  circles  touch  each  other  externally  and  thus 
inclose  one  acre  of  ground ;  find  the  radius  in  rods  of  each  of 
these  circles. 

11.  Show  that  in  two  circles  of  different  radii,  angles  at  the 
centres  subtended  by  arcs  of  equal  length  are  to  each  other  in- 
versely as  the  radii. 

12.  Show  that  the  square  on  the  side  of  a  regular  inscribed 
pentagon,  minus  the  square  on  the  side  of  a  regular  inscribed 
decagon,  is  equal  to  the  square  on  the  radius. 


224  GEOMETRY. BOOK    V. 


On  Constructions. 

Proposition  XIII.     Problem. 

387.    To  inscribe  a  regular  polygon  of  any  number  of 
sides  in  a  given  circle. 


Let  Q  be  the  given  circle,  and  n  the  number  of  sides 
of  the  polygon. 

It  is  required  to  inscribe  in  Q,  a  regular  polygon  having  n 
sides. 

Divide  the  circumference  of  the  O  into  n  equal  arcs. 

Join  the  extremities  of  these  arcs. 

Then  we  have  the  polygon  required. 

For  the  polygon  is  equilateral,  §  181 

{in  the  same  O  equal  aros  are  subtended  by  equal  chords) ; 

and  the  polygon  is  also  regular,  §  364 

{an  equilateral  polygon  inscribed  in  a  O  is  regular). 

Q.  E.  F 


CONSTRUCTIONS. 


225 


Proposition  XIV.     Problem. 

388.  To  inscribe  in  a  given  circle  a  regular  polygon 
which  has  double  the  number  of  sides  of  a  given  inscribed 
regular  polygon. 


Let  ABCD  be  the  given  inscribed  polygon. 

It  is  required  to  inscribe  a  regular  polygon  having  double  the 
number  of  sides  of ABC D. 

Bisect  the  arcs  A  B,  BC,  etc. 

Draw  AH,  E B,  B  F,  etc., 

The  polygon  AE  B  FC,  etc.,  is  the  polygon  required. 

For  the  chords  AB,  BC,  etc.,  are  equal,  §  363 

(being  sides  of  a  regular  polygon). 

.*.  the  arcs  AB,  BC,  etc.,  are  equal,  §  182 

(in  the  same  O  equal  chords  subtend  equal  arcs). 

Hence  the  halves  of  these  arcs  are  equal, 

or,  AE,  EB,  B F,  FC,  etc.,  are  equal j 

.'.  the  polygon  A  EB  F,  etc.,  is  equilateral. 

The  polygon  is  also  regular,  §  364 

(an  equilateral  polygon  inscribed  in  a  O  is  regular)  ; 

and  has  double  the  number  of  sides  of  the  given  regular 
polygon. 

Q.  E.  F. 


226 


GEOMETRY. BOOK  V. 


Proposition  XV.     Problem. 
389.  To  inscribe  a  square  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

It  is  required  to  inscribe  a  square  in  the  circle. 

Draw  the  two  diameters  A  C  and  B  D  _L  to  each  other. 

Join  AB,  BC,  CD,  and  DA. 

Then  A  B  C  D  is  the  square  required. 

For,  the  A  ABC,  BCD,  etc.,  are  rt.  A,  §  204 

(being  inscribed  in  a  semicircle) , 

and  the  sides  A  B,  B  C,  etc.,  are  equal,  §  181 

(in  the  same  O  equal  arcs  are  subtended  by  equal  chords)  ; 


.*.  the  figure  A  B  CD  is  a  square, 
(having  its  sides  equal  and  its  A  rt.  A  ). 


§  127 

Q.  E.  F. 


390.  Corollary.  By  bisecting  the  arcs  AB,  BC,  etc.,  a 
regular  polygon  of  8  sides  may  he  inscribed ;  and,  by  continuing 
the  process,  regular  polygons  of  16,  32,  64,  etc.,  sides  may  be 
inscribed. 


CONSTRUCTIONS.  227 


Proposition  XVI.     Problem. 
391.  To  inscribe  in  a  given  circle  a  regular  hexagon. 

/  /  V 

/  /      \ 

/  /        \ 


/j 

Let  0  be  the  centre  of  the  given  circle. 

It  is  required  to  inscribe  in  the  given  O  a  regular  hexagon. 

From  0  draw  any  radius,  as  0  0. 

From  G  as  a  centre,  with  a  radius  equal  to  0  G, 

describe  an  arc  intersecting  the  circumference  at  F. 

Draw  Oi^and  C F. 

Then  G  F  is  a  side  of  the  regular  hexagon  required. 

For  the  A  0  F  C  is  equilateral,  Cons. 

and  equiangular,  §  112 

.'.  the  Z  FO  0  is  J  of  2  rt.  A,  or,  J  of  4  rt.  A.        §  98 

.*.  the  arc  F C  is  \  of  the  circumference  ABC F, 

.''.  the  chord  FC,  which  subtends  the  arc  FG,  is  a  side 
of  a  regular  hexagon  ; 

and  the  figure  G  FD,  etc.,  formed  by  applying  the  radius 

six  times  as  a  chord,  is  the  hexagon  required. 

Q.  E.  F. 

392.  Corollary  1.  By  joining  the  alternate  vertices  A,  C, 
B,  an  equilateral  A  is  inscribed  in  a  circle. 

393.  Cor.  2.  By  bisecting  the  arcs  AB,  B  C,  etc.,  a  regu- 
lar polygon  of  12  sides  may  be  inscribed  in  a  circle;  and,  by 
continuing  the  process,  regular  polygons  of  24,  48,  etc.,  sides 
may  be  inscribed. 


228  GEOMETRY.  —  BOOK    V. 

Proposition  XYII.     Problem. 
394.   To  inscribe  in  a  given  circle  a  regular  decagon. 


Let  0  be  the  centre  of  the  given  circle. 

It  is  required  to  inscribe  in  the  given  O  a  regular  decagon. 

Draw  the  radius  0  G, 
and  divide  it  in  extreme  and  mean  ratio,  so  that  0  0  shall 
be  to  0  S  as  0  S  is  to  SG.  §311 

From  G  as  a  centre,  with  a  radius  equal  to  0  S, 
*        describe  an  arc  intersecting  the  circumference  at  B. 
Drawee,  £#,  and  BO. 
Then  B  G  is  a  side  of  the  regular  decagon  required. 
For  OG  :  OS  :  :  OS  :  SC,  Cons. 

and      •  BG=OS.  .  Cons. 

Substitute  for  0  S  its  equal  B  G, 
then  OG  :  BG  ::  BG  :  S  G. 

Moreover  the  Z  0  G  B  =  Z  S G  B,  Iden. 

.*.  the  A  OGB  and  B  GS  are  similar,  §  284 

(having  an  Z  of  the,  one  equal  to  an  Z.  of  the  other,  and  the  including  sides 
proportional). 

But  the  A  0  GB  is  isosceles,  §  160 

(its  sides  0  C  and  0  B  being  radii  of  the  same  circle). 

.•.the  A  B  G  S,  which  is  similar  to  the  A  0  GB,  is  isosceles, 


CONSTRUCTIONS.  229 


and                              BS  =  BC.  §114 

But                             OJS  =  BC,  Cons. 

.\OS  =  BS,  Ax.  1 
.*.  the  A  S  0  B  is  isosceles, 

and  theZ  0  =  ZSBOy  §112 

(being  opposite  equal  sides). 

But  the  Z  C  S  B  =  Z  0  +  Z  SB  0,  §  105 

(the  exterior  A  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A ). 

.\theZ  CSB  =  2Z  0. 
ZSCB(=Z  CSB)  =  2Z  0,  §112 

and  Z  OBC  (==  Z  SCB)  =  2  Z  0.  §112 

.'.  the  sum  of  the  A  of  the  A  0  0 B  =  5  Z  0. 

.\5  Z  0  =  2rt.  A,  §98 

and         Z  0  =  £  of  2  rt.  A,  or  ^  of  4  rt.  A 

.'.  the  arc  B  C  is  ^  of  the  circumference,  and 
.'.  the  chord  B  G  is  a  side  of  a  regular  inscribed  decagon. 

Hence,  to  inscribe  a  regular  decagon,  divide  the  radius  in 
extreme  and  mean  ratio,  and  apply  the  greater  segment  ten 
times  as  a  chord. 

Q.  E.  F. 

395.  Corollary  1.  By  joining  the  alternate  vertices  of  a 
regular  inscribed  decagon,  a  regular  pentagon  may  be  inscribed. 

396.  Cor  2.  By  bisecting  the  arcs  BC,  OF,  etc.,  a  regular 
polygon  of  20  sides  may  be  inscribed,  and,  by  continuing  the 
process,  regular  polygons  of  40,  80,  etc.,  sides  may  be  inscribed. 


230  GEOMETRY. BOOK   V. 


Proposition  XVIII.     Problem. 

397.  To  inscribe  in  a  given  circle  a  regular  pentedecagon, 
or  polygon  of  fifteen  sides. 


F 
Let  Q  be  the  given  circle. 
It  is  required  to  inscribe  in  Q  a  regular  pentedecagon. 
Draw  EH  equal  to  a  side  of  a  regular  inscribed  hexagon,    §  391 
and  E  F  equal  to  a  side  of  a  regular  inscribed  decagon.     §  394 

Join  FH. 
Then  FH  will  be  a  side  of  a  regular  inscribed  pentedecagon. 
For         the  arc  E  H  is  £  of  the  circumference, 

and  the  arc  E  F  is  ^  of  the  circumference ; 
.'.  the  arc  FH  is  £  —  ^  or  -fa,  of  the  circumference. 

.'.  the  chord  FH  is  a  side  of  a  regular  inscribed  pente- 
decagon, 

and  by  applying  FH  fifteen  times  as  a  chord,  we  have  the 
polygon  required. 

Q.  E.  F. 

398.  Corollary.  By  bisecting  the  arcs  FH,  HA,  etc., 
a  regular  polygon  of  30  sides  may  be  inscribed ;  and  by  con- 
tinuing the  process,  regular  polygons  of  60,  120,  etc.  sides  may 
be  inscribed. 


CONSTRUCTIONS.  231 


Proposition  XIX.  Problem. 

399.  To  inscribe  in  a  given  circle  a  regular  polygon 
similar  to  a  given  regular  polygon. 

cLr \#'  C  D 


Let  A  BCD,  etc.,  be  the  given  regular  polygon,  and 
C  D'  E  the  given  circle. 

It   is  required  to   inscribe   in    C  D'  E1   a  regular  polygon 
similar  to  A  B  G  D,  etc. 

From  0,  the  centre  of  the  polygon  ABC  D,  etc. 

draw  02)  and  0  C. 

From  0'  the  centre  of  the  O  C  D'  ®, 

draw  O1  C  and  O1  D', 

making  the  Z  0'  =  Z  0. 

Draw  CD'. 

Then  C  D'  will  be  a  side  of  the  regular  polygon  required. 

For   each   polygon  will  have  as  many  sides  as  the  Z.  0 
(=Z  0')  is  contained  times  in  4  rt.  A. 

.'.   the   potygon    C  D' E',   etc.    is    similar  to  the  polygon 
CDE,  etc.,  §372 

{two  regular  polygons  of  the  same  number  of  sides  are  similar). 

Q.  E.  F. 


232 


GEOMETRY. BOOK    V. 


Proposition  XX.     Problem. 

400.   To  circumscribe  about  a  circle  a  regular  polygon 
similar  to  a  given  inscribed  regular  polygon. 

BMC 


Let  HMRSy  etc.,  be  a  given  inscribed  regular  polygon. 

It   is  reqtrired  to  circumscribe  a  regular  polygon   similar 
to  HMRS,  etc. 

At  the  vertices  H,  M,  R,  etc.,  draw  tangents  to  the  O, 
intersecting  each  other  at  A,  B,  C,  etc. 

Then  the  polygon  ABC D,  etc.  will  be  the  regular  poly- 
gon required. 

Since  the  polygon  A  BC  D,  etc. 

has  the  same  number  of  sides  as  the  polygon  If  MRS,  etc., 

•it  is  only  necessary  to  prove  that  ABC D,  etc.  is  a  regular 
polygon.  §  372 


In  the  A  BHMfmd  C  M  R, 


HM=MR, 

(being  sides  of  a  regular  polygon). 


§  363 


CONSTBUCTIONS.  233 


the  A  BHM,  BMH,  C M R,  and  C  It  Mane  equal,     §  209 
(being  measured  by  halves  of  equal  arcs)  ; 

.-.  the  A  BHM  and  CM R  are  equal,  §  107 

(having  a  side  and  two  adjacent  A  of  the  one  equal  respectively  to  a  side  and 
two  adjacent  A  of  the  other). 

.'.ZB  =  ZC, 

(being  homologous  A  of  equal  A ). 

In  like  manner  we  may  prove  Z  C  =  Z  D,  etc 

.'.  the  polygon  ABC  D,  etc.,  is  equiangular. 

Since  the  A  BHM,  C  MR,  etc.  are  isosceles,       §  241 
(two  tangents  drawn  from  the  same  point  to  aO  are  equal), 

the  sides  B H,  B M,  CM,  C R,  etc.  are  equal, 
(being  homologous  sides  of  equal  isosceles  &  ). 

.'.the  sides  AB,  BC,  C  D,  etc.  are  equal,         Ax.  6 

and  the  polygon  ABC D,  etc.  is  equilateral. 

Therefore  the  circumscribed  polygon  is  regular  and  similar 
to  the  given  inscribed  polygon.  §  372 

Q.E  F. 


Ex.  Let  R  denote  the  radius  of  a  regular  inscribed  polygon, 
r  the  apothem,  a  one  side,  A  one  angle,  and  C  the  angle  at  the 
centre ;  show  that 

1.  In  a  regular  inscribed  triangle  a  =  R  ^3,  r  =  \  R, 
A  =00°,  C=  120°. 

2.  In  an  inscribed  square  a  =  R  \f2,  r  =  £  R  V2,  A  =  90°, 
C  =  90°. 

3.  In  a  regular  inscribed  hexagon  a  =  R,  r  =  J  R  tfS, 
^  =  120°,  (7  =  60°. 

R  (V^  -   1) 

4.  In    a    regular    inscribed    decagon     a   =  ^ > 

r  =  J  R  VlO  +  2  y/5,   ^  =  144°,    (7  =  36°. 


234 


GEOMETRY. 


BOOK    V. 


Proposition  XXI.     Problem. 

401.  To  find  the  value  of  the  chord  of  one-half  an  arc, 

in  terms  of  the  chord  of  the  whole  arc  and  the  radius  of  the 

circle. 

D 


Let  AB  be  the  chord  of  arc  A  B  and  A  D  the  chord 
of  one- half  the  arc  A  B. 

It  is  required  to  find  the  value  of  A  D  in  terms  of  A  B  and 
R  {radius). 

From  D  draw  D  H  through  the  centre  0, 

and  draw  0  A. 

HI)  is  A.  to  the  chord  A  B  at  its  middle  point  C,      §  60 

(two  points,  0  and  D,  equally  distant  from  the  extremities,  A  and  B,  de- 
termine the  position  of  a  A.  to  the  middle  point  of  A  B). 

The  Z  HAD  is  a  rt.  Z,  §  204 

(being  inscribed  in  a  semicircle), 

.\A~ff  =  DHX  DC,  §289 

(the  square  on  one  side  of  a  rt.  A  is  equal  to  the  product  of  the  hypotenuse  by 
the  adjacent  segment  made  by  the  ±  let  fall  from  the  vertex  of  the  rt.  Z  ). 

Now  DH=2R, 

and  J)C  =  DG-CO  =  B-CO; 

.\A~D2  =  2R(R-CO). 


CONSTRUCTIONS.  235 


Since  A  C  0  is  a  rt.  A, 

AO*  =  At?  +  CD2 ;  §331 

.-.  CO2  =  AO*  -  AC1. 


00  =  ^(AO2-AG2), 


=  SlR>-{\ABf, 
^S/lP-lJ-B*, 

»  4 

=  V4  R?  -  .Q2. 
2 

In  the  equation  if2?  =  2R  (R  —  CO), 

substitute  for  C  0  its  value  ^4  R2  ~  A  ° 


then  jTD2  =  2r(r-^EII), 

=  2  7?2  -  R  N±  R2  -  AB2\  . 

.'.AD  =  JiR?-rN±R?-  AB2\  . 

Q.  E.  F. 

402.  Corollary.    If  we  take  the  radius  equal  to  unity, 


the  equation  A  D  =  J2R2  -  R  N±R?-  IB2\  becomes 
AD  =  ^2-^i-ABi. 


236 


GEOMETRY. 


BOOK    V. 


Proposition  XXII.     Problem. 

403.   To   compute   the   ratio  of  the  circumference  of  a 
circle  to  its  diameter,  approximately. 


Since 


§376 


Let   C  be  the  circumference  and  R   the  radius  of  a 
circle. 

.    G 

C 
when  B  =  1,  ir  =  «-  • 

It  is  required  to  find  the  numerical  value  of  ir. 

We  make  the  following  computations  by  the  use  of  the 
formula  obtained  in  the  last  proposition, 


A  Z>  =  i/2  —  V4  -  A  B2, 


No. 
Sides. 

12 

24 

48 

96 

192 

384 

768 


when  A  B  is  a  side  of  a  regular  hexagon  : 
In  a  polygon  of 

Form  of  Computation. 

AD  —  VI 


Length  of  Side. 

V^-P .51763809 

AD  =  ^-\J±-  (.51 763809)2  .26105238 

AD  =  \J2-sJl-  (.26105238)2  .13080626 

A  D  =  sj'2  -  y/4  -  (.  1 3080626)2  .06543817 

A  D  =  \J2  -  y/4  -  (.06543817)2  .03272346 

A  D  =  V^2  —  y/4  —  (.03272346)2  .01636228 

AD  =  \j2-\l^-  (.01636228)2  .00818121 


Perimeter. 

6.21165708 
6.26525722 
6.27870041 
6.28206396 
6.28290510 
6.28311544 
6.28316941 


Hence  we  may  consider  6.28317  as  approximately  the  cir- 
cumference of  a  O  whose  radius  is  unity. 

. ,  ,          t    C          6.28317 
.  .  7r,  which  equals  —  ,  =  . 

2  2 


tt=  3.14159  nearly. 


Q.  E.  F 


ISOPERIMETRICAL    POLYGONS. 


237 


On  Isoperimetrical  Polygons.  —  Supplementary. 

404.  Def.  Isoperimetrical  figures  are  figures  which  have 
equal  perimeters. 

405.  Def.  Among  maguitudes  of  the  same  kind,  that 
which  is  greatest  is  a  Maximum,  and  that  which  is  smallest 
is  a  Minimum. 

Thus  the  diameter  of  a  circle  is  the  maximum  among  all 
inscribed  straight  lines;  and  a  perpendicular  is  the  minimum 
among  all  straight  lines  drawn  from  a  point  to  a  given  straight 
line. 


Proposition  XXIIL     Theorem. 

406.   Of  all  triangles  having  two  sides  respectively  equal, 
that  in  which  these  sides  include  a  right  angle  is  the  maxi- 


Let  the  triangles  ABC  and  EBC  have  the  sides  AB 
and  BG  equal  respectively  to  EB  and  BC;  and 
let  the  angle  ABC  be  a  right  angle. 

We  are  to  prove      A  ABO  A  EBC. 

From  E,  let  faU  the  ±  ED. 

The  A  ABC  and  EBC,  having  the  same  base  B C,  are  to 
each  other  as  their  altitudes  A  B  and  ED,  §  326 

(&  having  the  same  base  are  to  each  other  as  their  altitudes). 

Now  E  D  is  <EB,  §52 

(a  _L  is  the  shortest  distance  from  a  point  to  a  straight  line). 

But  EB  =  AB,  Hyp. 

..EDis<AB. 


A  ABO  A  EBC. 


Q.  E.  D. 


238  GEOMETRY. BOOK   V. 


Proposition  XXIV.     Theorem. 

407.  Of  all  polygons  formed  of  sides  all  given  but  one, 
the  polygon  inscribed  in  a  semicircle,  having  the  undetermined 
side  for  its  diameter,  is  the  maximum. 

C 


Let  AB,  BO,  0  D,  and  DE  be  the  sides  of  a  polygon 
inscribed  in  a  semicircle  having  A  E  for  its  di- 
ameter. 

We  are  to  prove  the  polygon  ABODE  the  maximum  of 
the  sides  A  B,  B  0,  0  D,  and  D  E. 


From  any  vertex,  as  0,  draw  0  A  and  C  E. 

Then  the  A  A  OE  is  a  rt.  Z. ,  §  204 

(being  inscribed  in  a  semicircle). 

Now  the  polygon  is  divided  into  three  parts,  ABO,  ODE, 
and  A  OE. 

The  parts  ABO  and  ODE  will  remain  the  same,  if  the 
Z  A  0  E  be  increased  or  diminished ; 

but  the  part  AGE  will  be  diminished,  §  406 

{of  all  &.  having  two  sides  respectively  equal,  that  in  which  these  sides  in- 
clude art.  A  is  the  maximum). 

.*.  A  B  0  D  E  is  the  maximum  polygon. 

Q.  E.  D. 


ISOPERIMETRICAL    POLYGONS. 


239 


Proposition  XXV.     Theorem. 

408.  The  maximum  of  all  polygons  formed  of  given  sides 
can  be  inscribed  in  a  circle. 

W      D 


A  A' 

Let  ABODE  be  a  polygon  inscribed  in  a  circle,  and 
A' B' C  D' E'  be  a  polygon,  equilateral  with  re- 
spect to  ABODE,  but  which  cannot  be  inscribed 
in  a  circle. 

We  are  to  prove 

the  polygon  A  B  0  D  E  >  the  polygon  A'  B'  C  D'  E'. 

Draw  the  diameter  A  H. 

Join  OH  and  D  H. 

Upon  C  D>  (=  0  D)  construct  the  A  C  H'  D1  =  A  C  HD, 

and  draw  A'  W. 

Now  the  polygon  A  B  0  H  >  the  polygon  A'  B'  C  H',     §  407 

(of  all  polygons  formed  of  sides  all  given  but  one,  the  polygon  inscribed  in  a 
semicircle  having  the  nndetermhied  side  for  its  diameter,  is  the  maximum). 

And  the  polygon  A  E  D  H  >  the  polygon  A'  E'  D'  H'.     §  407 

Add  these  two  inequalities,  then 

the  polygon  A  BO  HDE>  the  polygon  A' B'  C'H'D'E'. 

Take  away  from  the  two  figures  the  equal  A  0  H  D  and 
C'H'D'. 

Then  the  polygon  A  B  0  D  E  >  the  polygon  A'  B'  C  D'  E'. 

Q.  E. O, 


240 


GEOMETRY. BOOK    V. 


Proposition  XXVI.     Theorem. 

409.   Of  all  triangles  having  the  same  base  and  equal 
perimeters,  the  isosceles  triangle  is  the  maximum. 


^H 


Let   the  AACB  and  ABB  have   equal  perimeters, 
and  let  the  A  AC B  be  isosceles. 

We  are  to  prove      AAOB>AADB. 

Draw  the  Js  CE  and  B  F. 


A  A  OB 


CE 
BF 


A  ABB 

(A  having  the  same  base  are  to  each  other  as  their  altitudes). 

Produce  AC  to  ff,  making  C  ff=  AC. 
Draw  H B. 


§  326 


The  Z  A  B  H  is  a  rt.  Z,  for  it  will  be  inscribed  in  the 
semicircle  drawn  from  C  as  a  centre,  with  the  radius  C  B< 


ISOPERIMETEICAL   POLYGONS.  241 


From  C  let  fall  the  X  C  K ; 
and  from  D  as  a  centre,  with  a  radius  equal  to  D  Bf 
describe  an  arc  cutting  H  B  produced,  at  P. 
Draw  DP  and  A  P, 
and  let  fall  the  ±  D  M. 
Since        AH  =  AC+CB  =  AD  +  DB, 
and  AP<AD+  DP; 

.:AP<AD  +  DB; 
.\AH>AP. 
.\BH>BP.  §56 

Now  BK=%BH,  §113 

(a  _L  drawn  from  the  vertex  of  an  isosceles  A  bisects  the  base), 

and  BM=iBP.  §113 

But  CE  =  BK,  §135 

(lb  comprehended  between  lis  are  equal); 

and  DF=BM,  §136 

.-.  CE>  DF. 
.\AACB>AADB. 

Q.  E.  D. 


242 


GEOMETRY.  —  BOOK    V. 


Proposition  XXVII.     Theorem. 

410.   The  maximum  of  isoperimetrical  polygons  of  the 
same  number  of  sides  is  equilateral. 


Let  ABC  D,  etc.,  be  the  maximum  of  isoperimetrical 
polygons  of  any  given  number  of  sides. 

We  are  to  prove      AB,  BC,  CD,  etc.,  equal. 
Draw  A  G. 

The  A  AB C  must  be  the  maximum  of  all  the  A  which 
are  formed  upon  A  G  with  a  perimeter  equal  to  that  of  A  ABG. 

Otherwise,  a  greater  A  A  KG  could  be  substituted  for  A  A  B  G, 
without  changing  the  perimeter  of  the  polygon. 

But  this  is  inconsistent  with  the  hypothesis  that  the  poly- 
gon ABC  D,  etc.,  is  the  maximum  polygon. 

.*.  the  A  A  B  G,  is  isosceles,  §  409 

(of  all  &  having  the  same  base  and  equal  perimeters,  the  isosceles  A  is  the 
maximum). 

In  like  manner  it  may  be  proved  that  B  C  =  CD,  etc. 

Q.  E.  D. 

411.  Corollary.    The   maximum  of  isoperimetrical  poly- 
gons of  the  same  number  of  sides  is  a  regular  polygon. 

For,  it  is  equilateral,  §  410 

(the  maximum  of  isoperimetrical  polygons  of  the  same  number  of  sides  is 
equilateral). 

Also  it  can  be  inscribed  in  a  O,  §  408 

(the  maximum  of  all  polygons  formed  of  given  sides  can  be  inscribed  in  a  O). 

Hence  it  is  regular,  §  364 

(an  equilateral  polygon  inscribed  in  a  O  is  regular). 


ISOPERIMETRICAL    POLYGONS. 


243 


Proposition  XXVIII.     Theorem. 

412.   Of  isoperimetrica I  regular  polygons,  tha t  is  greates t 
which  has  the  greatest  number  of  sides. 


Let  Q  be  a  regular  polygon  of  three  sides,  and  Q'  be 
a  regular  polygon  of  four  sides,  each  having  the 
same  perimeter. 

We  are  to  prove      Q'  >  Q. 

In  any  side  A  B  of  Q,  take  any  point  D. 

The  polygon  Q  may  be  considered  an  irregular  polygon 
of  four  sides,  in  which  the  sides  A  D  and  D  B  make  with  each 
other  an  Z  equal  to  two  rt.  A . 

Then  the  irregular  polygon  Q,  of  four  sides  is  less  than  the 

regular  isoperi  metrical  polygon  Q'  of  four  sides,  §  411 

(the  maximum  of  isoperimetrical  polygons  of  the  same  number  of  sides  is  a 
regular  polygon). 

In  like  manner  it  may  be  shown  that  Qf  is  less  than  a 
regular  isoperimetrical  polygon  of  five  sides,  and  so  on. 

Q.  E.  D. 


413.  Corollary.    Of  all  isoperimetrical  plane  figures  the 
circle  is  the  maximum. 


244 


GEOMETRY.  —  BOOK   V. 


Proposition  XXIX.     Theorem. 

414.  If  a  regular  polygon  be  constructed  with  a  given 
area,  its  perimeter  will  be  the  less  the  greater  the  number 
of  its  sides. 


Let  Q  and  Q'  be  regular  polygons  having  the  same 
area,  and  let  Q'  have  the  greater  number  of  sides. 

We  are  to  prove  the  perimeter  of  Q  >  the  perimeter  of  Ql. 

Let  Q"  be  a  regular  polygon  having  the  same  perimeter  as 
Q',  and  the  same  number  of  sides  as  Q. 

Then  Q  is  >  Q",  §  412 

{of  isoperimetrical  regular  polygons,  that  is  the  greatest  which  has  the  greatest 
number  of  sides). 

But  Q  =  Q', 

.-.Qis>  Q". 
.\  the  perimeter  of  Q  is  >  the  perimeter  of  Q". 
But  the  perimeter  of  Q'  =  the  perimeter  of  Q",      Cons. 
.*.  the  perimeter  of  Q  is  >  that  of  Q'. 

Q.  E.  D. 

415.  Corollary.   The  circumference  of  a  circle  is  less  than 
the  perimeter  of  any  other  plane  figure  of  equal  area. 


SYMMETRY. 


245 


On  Symmetry.  —  Supplementary. 

416.  Two  points  are  Symmetrical  when  they  are  situated 
on  opposite  sides  of,  and  at  equal  distances  from,  a  fixed  point, 
line,  or  plane,  taken  as  an  object  of  reference. 

417.  When  a  point  is  taken  as  an  object  of  reference,  it  is 
called  the  Centre  of  Symmetry ;  when  a  line  is  taken,  it  is  called 
the  Axis  of  Symmetry ;  when  a  plane  is  taken,  it  is  called  the 
Plane  of  Symmetry. 


418.  Two  points  are  symmetrical  with  re- 
spect to  a  centre,  if  the  centre  bisect  the  straight 
line  terminated  by  these  points.  Thus,  P,  P' 
are  symmetrical  with  respect  to  C,  if  C  bisect 
the  straight  line  PP. 


419.  The  distance  of  either  of  the  two  symmetrical  points 
from  the  centre  of  symmetry  is  called  the  Radius  of  Symmetry. 
Thus  either  C  P  or  C  P'  is  the  radius  of  symmetry. 

420.  Two  points   are    symmetrical   with  P 
respect  to  an  axis,  if  the  axis  bisect  at  right 

angles  the   straight  line  terminated  by  these   X X? 

points.     Thus,  P,  P'  are  symmetrical  with  re- 
spect to  the  axis  XX',  if  XX'  bisect  P  P'  at  p, 
right  angles. 


421.  Two  points  are  symmetrical  with 
respect  to  a  plane,  if  the  plane  bisect  at 
right  angles  the  straight  line  terminated  by 
these  points.  Thus  P,  P'  are  symmetrical 
with  respect  to  M  N,  if  M  N  bisect  P  P'  at 
right  angles. 


W 


P> 


N 


246 


GEOMETRY. BOOK  V. 


422.  Two  plane  figures  are  symmetrical  with  respect  to  a 
centre,  an  axis,  or  a  plane,  if  every  point  of  either  figure  have 
its  corresponding  symmetrical  point  in  the  other. 

A 


A' 


Fig.  2. 


Fig.  3. 


Thus,  the  lines  A  B  and  A'  B'  are  symmetrical  with  respect 
to  the  centre  G  (Fig.  1),  to  the  axis  XX  (Fig.  2),  to  the  plane 
M  N  (Fig.  3),  if  every  point  of  either  have  its  corresponding 
symmetrical  point  in  the  other. 


\  ' 

Z) 

\ 

*'■/ 

/! 

\ 

1 

Ml— 

I 

1 

V 

j 

/b\ 

N 


A'  D' 

Fig.  6. 

Also,  the  triangles  ABB  and  A1  B'  D'  are  symmetrical  with 
respect  to  the  centre  C  (Fig.  4),  to  the  axis  XX'  (Fig.  5),  to  the 
plane  MN  (Fig.  6),  if  every  point  in  the  perimeter  of  either 
have  its  corresponding  symmetrical  point  in  the  perimeter  of  the 
other. 

423.  Def.  In  two  symmetrical  figures  the  corresponding 
symmetrical  points  and  lines  are  called  homologous. 


SYMMETRY. 


247 


Two  symmetrical  figures  with  respect  to  a  centre  can  be 
brought  into  coincidence  by  revolving  one  of  them  in  its  own 
plane  about  the  centre,  every  radius  of  symmetry  revolving 
through  two  right  angles  at  the  same  time. 

Two  symmetrical  figures  with  respect  to  an  axis  can  be 
brought  into  coincidence  by  the  revolution  of  either  about  the 
axis  until  it  comes  into  the  plane  of  the  other. 

424.  Def.  A  single  figure  is  a  symmetrical  figure,  either 
when  it  can  be  divided  by  an  axis,  or  plane,  into  two  figures 
symmetrical  with  respect  to  that  axis  or  plane ;  or,  when  it  has 
a  centre  such  that  every  straight  line  drawn  through  it  cuts  the 
perimeter  of  the  figure  in  two  points  which  are  symmetrical 
with  respect  to  that  centre. 


Fig.  1. 


Fig.  2. 


Thus,  Fig.  1  is  a  symmetrical  figure  with  respect  to  the 
axis  XX't  if  divided  by  XX'  into  figures  ABC D  and  AB'C'D 
which  are  symmetrical  with  respect  to  XX'. 

And,  Fig.  2  is  a  symmetrical  figure  with  respect  to  the 
centre  0,  if  the  centre  0  bisect  every  straight  line  drawn 
through  it  and  terminated  by  the  perimeter. 

Every  such  straight  line  is  called  a  diameter. 

The  circle  is  an  illustration  of  a  single  figure  symmetrical 
with  respect  to  its  centre  as  the  centre  of  symmetry,  or  to  any 
diameter  as  the  axis  of  symmetry. 


248  GEOMETRY.  —  BOOK   V. 


Proposition  XXX.     Theorem. 

425.  Two  equal  and  parallel  lines  are  symmetrical  with 
respect  to  a  centre. 

A  B> 


B  A' 

Let  A  B  and  A1 B'  be  equal  and  parallel  lines. 
We  are  to  prove      A  B  and  A'  B'  symmetrical. 
Draw  A  A'  and  B  &t  and  through  the  point  of  their  inter- 
section G,  draw  any  other  line  EG H',  terminated  in  A  B  and 
A'B'. 

In  the  A  GABzxAGA'B' 

AB  =  A'B',  Hyp. 

also,     A  A  and  B  =  A  A'  and  B'  respectively,  §  68 

(being  alt.  -int.  A  ), 

.'.AGAB  =  A  GA'B';  §  107 

.'.  GA  and  G  B  =  G  A'  and  G B'  respectively, 
(being  homologous  sides  of  equal  &). 

Now  in  the  A  A  G  R  and  A' C  H' 

AC  =  A'G, 

A  A  and  AC  H  —  A  A'  and  A'  G II1  respectively, 

.    .'.AAGH^AA'GH',  §  107 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other). 

.'.GH^CH', 

(being  homologous  sides  of  equal  A  ). 

.*•  EP  is  the  symmetrical  point  of  H. 

But  H  is  any  point  in  A  B ; 

.'.  every  point  in  A  B  has  its  symmetrical  point  in  A' B1. 

.'.  A  B  and  A!  B'  are  symmetrical  with  respect  to  G  as  a 
centre  of  symmetry. 

Q.  E.  D. 

426.  Corollary.  If  the  extremities  of  one  line  be  re- 
spectively the  symmetricals  of  another  line  with  respect  to  the 
same  centre,  the  two  lines  are  symmetrical  with  respect  to  that 
centre. 


SYMMETRY. 


249 


Proposition  XXXI.     Theorem. 

427.  If  a  figure  be  symmetrical  with  respect  to  two  axes 
'perpendicular  to  each  other,  it  is  symmetrical  with  respect 
to  their  intersection  as  a  centre. 


Let  the  figure  ABODE FGH  be  symmetrical   to    the 

two  axes  XX',  YY'  which  intersect  at  0. 

We  are  to  prove  0  the  centre  of  symmetry  of  the  figure. 

Let  /  be  any  point  in  the  perimeter  of  the  figure. 

Draw  IKL  ±  to  XX',  and  I M N  J_  to  YY. 

JoinZO,  ON,  and  KM. 

Now  KI=KL, 

(tJie  figure  being  symmetrical  with  respect  to  X  XI). 

But  KI=OM, 

(lis  comprehended  between  lis  are  egual). 

,'.KL  =  OM. 

.\  K  LO  M  is  a  O, 

{having  two  sides  equal  and  parallel). 

.'.  LO  is  equal  and  parallel  to  KM, 
{being  opposite  sides  of  a  O). 

In  like  manner  we  may  prove  0  N  equal  and  parallel  to  KM. 
Hence  the  points  L,  0,  and  N  are  in  the  same  straight  line 
drawn  through  the  point  0  11  to  KM. 

Also  L0  =  0N, 

(since  each  is  equal  to  KM). 
*,  any  straight  line  LO  N,  drawn  through  0,  is  bisected  at  0. 
.'.  0  is  the  centre  of  symmetry  of  the  figure.        §  424 

Q.  E.  D. 


§420 

§  135 

Ax.  1 
§  136 

§  134 


250 


GEOMETRY.  —  BOOK    V. 


Exercises. 
1.   The  area  of  any  triangle  may  be  found  as  follows  :  From 
half  the  sum  of  the  three  sides  subtract  each  side  severally,  mul- 
tiply together  the  half  sum  and  the  three  remainders,  and  extract 
the  square  root  of  the  product. 

Denote  the  sides  of  the  tri- 
angle A  B  G  by  a,  b,  c,  the  alti- 
a+  6  +  c 


tude  by  p,  and 


by  s. 


Show  that 


,2  = 


+  c2-2cXAD, 


and  show  that 
P2  =  b2- 


(b2+<?-a2y 
4  c2 


p  = 


p  = 


V/462c2-(62+c2-a2)2 


2c 


\/  (b  +  c+  a)  (b  +  c—  a)  (a  +  b  -  c)  (a—  b  +  c) 


2c 


Hence,  show  that  area  of  A  A  B  C,  which  is  equal  to 


cXp 


=  i^  (b  +  c  +  a)  (b  +  c-a)  (a  +  b-c)  (a-b  +  c), 
=  \/  s(s  —  a)(s  —  b)(s  —  c). 

2.  Show  that  the  area  of  an  equilateral  triangle,  each  side  of 
which  is  denoted  by  a,  is  equal  to  — j-  . 

3.  How  many  acres  are  contained  in  a  triangle  whose  sides 
are  respectively  60,  70,  and  80  chains  1 

4.  How  many  feet  are  contained  in  a  triangle  each  side  of 
which  is  75  feet  1 


BOOK   VI. 

PLANES  AND  SOLID  ANGLES. 


On  Lines  and  Planes. 


428.  Def.  A  Plane  has  already  been  denned  as  a  surface 
such  that  the  straight  line  joining  any  two  points  in  it  lies 
wholly  in  the  surface. 

The  plane  is  considered  to  be  indefinite  in  extent,  so  that 
however  far  the  straight  line  be  produced,  all  its  points  lie  in 
the  plane.  A  plane  is  usually  represented  by  a  quadrilateral 
supposed  to  lie  in  the  plane. 

429.  Def.  The  Foot  of  a  line  is  the  point  in  which  it 
meets  the  plane. 

430.  Def.  A  straight  line  is  perpendicular  to  a  plane  if 
it  be  perpendicular  to  every  straight  line  of  the  plane  drawn 
through  its  foot. 

In  this  case  the  plane  is  perpendicular  to  the  line. 

431.  Def.  The  Distance  from  a  point  to  a  plane  is  the 
perpendicular  distance  from  the  point  to  the  plane. 

432.  Def.  A  line  is  parallel  to  a  plane  if  all  its  points  be 
equally  distant  from  the  plane. 

In  this  case  the  plane  is  parallel  to  the  line. 

433.  Def.  A  line  is  oblique  to  a  plane  if  it  be  neither  per- 
pendicular nor  parallel  to  the  plane. 

434.  Def.  Two  planes  are  parallel  if  all  the  points  of 
either  be  equally  distant  from  the  other. 

435.  Def.  The  Projection  of  a  point  on  a  plane  is  the  foot 
of  the  perpendicular  from  the  point  to  the  plane. 

436.  Def.  The  projection  of  a  line  on  a  plane  is  the  locus 
of  the  projections  of  all  its  points. 

437.  Def.  The  plane  embracing  the  perpendiculars  which 
project  the  points  of  a  straight  line  upon  a  plane  is  called  the 
projecting  plane  of  the  line. 


J 


252  GEOMETRY.  —  BOOK   VI. 

438.  Def.  The  angle  which  a  line  makes  with  a  plane  is 
the  angle  which  it  makes  with  its  projection  on  the  plane. 

This  angle  is  called  the  Inclination  of  the  line  to  the  plane. 

439.  Def.  A  plane  is  determined  by  lines  or  points,  if 
no  other  plane  can  embrace  these  lines  or  points  without  being 
coincident  with  that  plane. 

440.  -Def.  The  intersection  of  two  planes  is  the  locus  of 
all  the  points  common  to  the  two  planes. 

441.  An  infinite  number  of  planes  may  embrace  the  same 
straight  line. 

Thus,  if  the  plane  M  N  em-  M_ 

brace  the  line  AB  it  may  be  made 
to  revolve  about  A  B  as  an  axis, 

and  to  occupy  an  infinite  number    ^  ^^ /f 

of  positions,  each  of  which  is  the         /  / 

position  of  a  plane  embracing  the 
line  A  B. 

442.  A  plane  is  determined  by  a  straight  line  and  a  point 
without  that  line. 

Thus,  let  any  plane  em- 
bracing the  straight  line  A  B 
revolve  about  the  line  as  an  axis 
until  it  embraces  the  point  C. 

Now  if  the  plane  revolve  either  way  about  the  line  A  B  as 
an  axis,  it  will  cease  to  embrace  the  point  G. 

Hence  any  other  plane  embracing  the  line  A  B  and  the 
point  C  must  be  coincident  with  the  first  plaie.  §  439 

443.  Three  points  not  in  a  straight  line  determine  a  p>lane. 
For,  by  joining  any  two  of  the  points ,  we  have  a  straight 

line  and  a  point  which  determine  a  plane.  §  442 

444.  Two  intersecting  straight  lines  determine  a  plane. 

For,  a  plane  embracing  one  of  these  straight  lines  and  any 
point  of  the  other  line  (except  the  point  of  intersection)  is  deter- 
mined. §  442 

445.  Two  parallel  straight  lines  determine  a  plane. 

For,  a  plane  embracing  either  of  these  parallels  and  any 
point  in  the  other  is  determined.  §  442 


LINES    AND   PLANES. 


253 


Proposition  I.     Theorem. 

446.   If  two  planes  cut  one  another  their  intersection  is 
a  straight  line. 


Let  MN  and  PQ  be  two  planes  which  cut  one  another. 

We  are  to  prove  their  intersection  a  straight  line. 

Let  A  and  B  be  two  points  common  to  the  two  planes. 

Draw  the  straight  lino  A  B. 

Since  the  points  A  and  B  are  common  to  the  two  planes, 
the  straight  line  A  B  lies  in  both  planes.  §  428 

Now,  no  point  out  of  this  line  can  be  in  both  planes ; 

for,  if  it  be  possible,  let  C  be  such  a  point. 
But  there  can  be  but  one  plane  embracing  the  point  0  and 
the  line  A  B.  §  442 

.'.  C  does  not  lie  in  both  planes. 

.'.  every  point  in  the  intersection  of  the  two  planes  lies  in 
the  straight  line  A  B. 

Q.  E.  D 


254 


GEOMETRY.  —  BOOK  VI. 


Proposition  II.     Theorem. 

447.  From  a  point  without  a  plane  only  one  perpendic- 
ular can  be  drawn  to  the  plane ;  and  at  a  given  point  in  a 
plane  only  one  perpendicular  can  be  erected  to  the  plane. 


IV 


7 


Fig.  l. 


N 


Fig.  2. 


Let  G  D    (Fig.l)  be  a,  perpendicular  let  fall  from  the 
point  G  to  the  plane  MN. 

We  are  to  prove  that  no  other  J_  can  be  drawn  from  the  point 
C  to  the  plane  MN 

If  it  be  possible,  let  G  B  be  another  _L  to  the  plane  MN, 

and  let  a  plane  P  Q  pass  through  the  lines  G  B  and  0  D. 

The  intersection  of  P  Q  with  the  plane  MN  is  a  straight 
line  BD.  §446 

Now  if  CD  and  GB  be  both  J_  to  the  plane,  the  A  GBD 
would  have  two  rt.  A,  GBD  and  G D B,  which  is  impos- 
sible. §  102 

Let  D  G  {Fig.  2)  be  a  perpendicular  to  the  plane  MN  at 
the  point  D. 

If  it  be  possible,  let  D  A  be  another  JL  to  the  plane  from 
the  point  D} 

and  let  a  plane  P Q  pass  through  the  lines  D  G  and  DA. 

The  intersection  of  P  Q  with  the  plane  JOris  a  straight  line. 

Now  if  D  G  and  DA  could  both  be  J_  to  the  plane  MN  at 
D,  we  should  have  in  the  plane  P  Q  two  straight  lines  _L  to  the 
line  D  Q  at  the  point  D,  which  is  impossible.  §  61 

Q.  E.   D. 

448.  Corollary.  A  perpendicular  is  the  shortest  distance 
from  a  point  to  a  plane. 


LINES    AND    PLANES. 


255 


Proposition  III.     Theorem. 

449.  If  a  straight  line  be  perpendicular  to  each  of  two 
straight  lines  drawn  through  its  foot  in  a  plane  it  is  perpen. 
dicular  to  the  plane. 


Let  DC  be  perpendicular  to  each  of  the  two  lines 
AC  A'  and  BCB'  drawn  through  its  foot  in  the 
plane  M  N. 

We  are  to  prove  DC  A- to  the  plane  MN. 

Take  CA  =  C  A'  and  C  B  =  CB'. 

JomABttidiA'B'. 

Then  A  B  and  A'  B'  are  symmetrical  with  respect  to  C,  §  426 
{their  extremities  being  symmetrical). 
Through  C  draw  any  line  HC  H'  in  the  plane  M  N. 

Then  H  and  H'  are  symmetrical,  §  422 

(being  corresponding  points  in  the  symmetrical  lines  A  B  and  A1 2?'). 

About  C,  the  centre  of  symmetry,  revolve  A  B,  keeping  A  C 
and  B  C  _L  to  CD,  until  it  comes  into  coincidence  with  A' B'. 

Then  the  point  H  will  coincide  with  its  symmetrical 
point  H\ 

and  Z  DCH  will  coincide  with,  and  be  equal  to,  Z  DCH1. 

.*.  A  DCHandi  DCW  are  rt.  A.  §  25 

.'.DC is  ±  to  HCH'. 

Now  since  DC  is  _L  to  any  line,  HCH\  drawn  through 
its  foot  in  the  plane  MN,  it  is  _L  to  every  such  line. 

§430. 
a  e.  d. 


\  DC  is  i_  to  the  plane  MN. 


256 


GEOMETRY. BOOK   VI. 


Proposition  IV.     Theorem. 
450.    Oblique  lines  drawn  from  a  point  to  a  plane  at 
equal  distances  from  the  foot  of  the  perpendicular  are  equal; 
and  of  two  oblique  lines  unequally  distant  from  the  foot  cfthe 
perpendicular  the  more  remote  is  the  greater. 


Let  the  oblique  lines  BC,  B D,  and  BE,  be  drawn  at 
equal  distances,  AC,  AD,  and  A  E,  from  the  foot 
of  the  perpendicular  BA  ;  and  let  BC  be  drawn 
more  remote  from  the  foot  of  the  perpendicular 
than  BC. 

We  are  to  prove   I.    BC=BD  =  BE. 
II.   BOBC. 

I.  In  the  rt.  ABAC  and  BA  D 

BA=BA, 
AC  =  AD, 
and  rt.  A  BA  C  =  rt.  Z  BA  D. 

.'.ABAC  =  ABAZ>, 

,\BC  =  BD, 

(being  homologous  sides  of  equal  &). 

II.  Since  A  C  is  >  A  C, 

BC'is>BC, 


Iden. 
Hyp. 

§106 


§55 

Q.  E.  6. 

451.  Cor.  1.  Equal  oblique  lines  from  a  point  to  a  plane 
meet  the  plane  at  equal  distances  from  the  foot  of  the  perpendic- 
ular; and  of  two  unequal  oblique  lines,  the  greatermeets  the  plane 
at  the  greater  distance  from  the  foot  of  the  perpendicular. 

452.  Cor.  2.  All  equal  oblique  lines  BC,  B D,  etc.,  drawn 
from  a  point  to  a  plane  terminate  in  the  circumference  CDE 
described  from  A  as  a  centre  with  a  radius  equal  to  A  C.  Hence, 
to  draw  a  perpendicular  from  a  point  to  a  plane,  draw  any  ob- 
lique line  from  the  given  point  to  the  plane ;  revolve  this  line 
about  the  point,  tracing  the  circumference  of  a  circle  in  the  plane, 
and  draw  a  line  from  the  point  to  the  centre  of  the  circle. 


LINES    AND    PLANES.  257 

Proposition  Y.     Theorem. 

453.  If  three  straight  lines  meet  at  one  pointy  and  a 
straight  line  be  perpendicular  to  each  of  them  at  that  point, 
the  three  straight  lines  lie  in  the  same  plane. 


P:"  A 


Let  the  straight  line  A  B  be  perpendicular  to  each  of 
the  straight  lines  BC,  BD,  and  B  E,  at  B. 

We  are  to  prove  B  C,  B  D,  and  BE  in  the  same  plan-e  M N. 

If  not,  let  B  D  and  BE  be  in  the  plane  M  N,  and  BC  with- 
out it ;  and  let  P  H,  passing  through  A  B  and  B  C,  cut  the  plane 
M  N  in  the  straight  line  B  H. 

Now      A  B,  BC,  and  B U are  all  in  the  plane  P H, 
and  since   A  B  is  _L   to  B  D  and  B  E,  it  is  _L  to  the 
plane  M  N,  §  449 

(if  a  straight  line  be  _L  to  each  of  two  straight  lines  drawn  through  its  foot 
in  a  plane,  it  is  ±  to  tlie  plane). 

.'.  A  B  is  _L  to  B II,  a  straight  line  in  the  plane  M N,  §  430 

(a  _L  to  a  plane  is  1.  to  every  straight  line  in  that  plane  drawn  through 

Us  foot). 

That  is  Z  ABU  is  art.  Z. 

But  Z  ABC  is  a  rt.  Z.  Hyp. 

r.Z  ABC  =  Z  ABU. 

.'.  BC  and  B1I  coincide. 

.*.  B  C  is  not  without  the  plane  M N. 

Q.  E.  D 

454.  Corollary.  The  locus  of  all  perpendiculars  to  a  given 
straight  line  at  a  given  point  is  a  plane  perpendicular  to  this 
given  straight  line  at  the  given  point. 

455.  Scholium.  In  the  geometry  of  space  the  term  locus 
has  the  same  signification  as  in  plane  geometry,  only  it  is  not 
limited  to  lines,  but  is  extended  to  include  surfaces. 


258 


GEOMETRY. BOOK  VI. 


Proposition  VI.     Theorem. 

456.  If  from  the  foot  of  a  perpendicular  to  a  plane  a 
straight  line  be  drawn  at  right  angles  to  any  line  of  the  plane, 
the  line  drawn  from  its  intersection  with  the  line  of  the  plane 
to  any  point  of  the  perpendicular  is  perpendicular  to  the  line 
of  the  plane. 


Let  P  F  be  a  perpendicular  to  the  plane  M N,  FC 
a  perpendicular  from  the  foot  of  P F  to  any  line 
AB,  in  the  plane  M ' N,  and  CP  a  line  drawn  from 
its  intersection  with  A  B  to  any  point  P  in  the 
perpendicular  P  F. 

We  are  to  prove  G  P  -L  to  A  B. 

Take  GA  =  CB  and  draw  FA,  FB,  P  A,  P B. 

Now  FA  =  FB,  §  53 

(two  oblique  lines  drawn  from  a  point  in  a  X  cutting  off  equal  distances 
from  the  foot  of  the  ±  are  equal), 

and  PA=  PB,  §450 

(oblique  lines  drawn  from  a  point  to  a  plane  at  equal  distances  from  the 
foot  of  the  J-  are  equal). 

.'.  PG  is  ±to  A  B,  §60 

(two  points  equally  distant  from  the  extremities  of  a  straight  line  determine 
the  X  at  the  middle  point  of  the  line). 

Q.  E.  D 


LINES    AND    PLANES. 


259 


Proposition  VII.     Theorem. 

457.  If  a  line  be  perpendicular  to  a  plane,  every  line 
which  is  parallel  to  this  perpendicular  is  likewise  perpendic- 
ular to  the  plane. 

A  C 


/J 


^\7F 7 

ML n 


Let  AB  be  perpendicular  to  the  plane  M N,  and  CD 

any  line  parallel  to  AB. 

We  are  to  prove  C  D  perpendicular  to  the  plane  M  N. 

Draw  B  D  in  the  plane  M  N,  and  through  D  draw  E  F  in 
the  plane  M N  _L  to  B  D,  and  join  D  with  any  point  in  A  B, 
as  A. 

BDis±toAB,  §430 

(if  a  straight  line  be  ±  to  a  plane  it  is  ±  to  every  line  of  tJie  plane  drawn 
through  its  foot)  ; 

itisalso_LtoCZ),  §67 

(if  a  straight  line  be  X  to  one  of  two  lis,  it  is  A.  to  the  oilier). 

Now  EF is  X  to  AD,  §  456 

(if  from  the  foot  of  a  JL  to  a  plane  a  straight  line  be  drawn  at  right  angles  to 

any  line  of  the  plane,  the  line  drawn  from  its  intersection  with  the  line 

of  the  plane  to  any  point  in  the  ±  is  ±  to  the  line  of  the  plane), 

and  is  also  _L  to  B  D.  Cons. 

.-.  E F is  _L  to  the  plane  ABDC,  §  449 

(a  straight  line  X  to  two  straight  lines  drawn  through  its  foot  in  a  plane  is 
JL  to  the  plane), 

.\EFis±to  CD,  §  430 

(if  a  straight  line  be  J-  to  a  plane  it  is  _L  to  every  line  of  the  plane  drawn 
through  its  foot). 

.'.  CD  is  _L  to  BD  and  EF,  and  consequently  to  the 
plane  MN.  §  449 

Q.  E.  D. 

458.  Corollary  1.  Two  lines  which  are  perpendicular  to 
the  same  plane  are  parallel. 

459.  Cor.  2.  Two  lines  parallel  to  a  third  straight  line  not 
in  their  own  plane  are  parallel  to  each  other. 


260 


GEOMETRY.  —  BOOK  VI. 


Proposition  VIII.     Theorem. 

460.    If  a  straight  line  and  a  plane  be  perpendicular  to 
the  same  straight  line,  they  are  parallel. 


Let  the  straight  line  B  G  and  the  plane  M N  be  per- 
pendicular to  the  straight  line  A  B. 

We  are  to  prove         B  C  II  to  M  N. 

From  any  point  G  of  the  line  BC  let  G  D  be  drawn  per- 
pendicular to  M  N. 

Join  A  D. 

B  A  and  C  D  are  parallel,  §  458 

(two  straight  lines  _L  to  the  same  plane  are  II ). 

ADisLtoBA,  §430 

(if  a  straight  line  be  ±  to  a  plane  it  is  _L  to  every  line  of  the  plane  drawn 
through  its  foot). 

.*.  A  D  and  B  C  are  parallel,  §  65 

(two  straight  lines  JL  to  the  same  straight  line  are  II ). 

.-.  A  BCD  is  aO.  §125 

.\CD  =  AB.  §134 

Now,  since  G  is  any  point  in  the  line  B  G,  all  the  points  in 
B  G  are  equally  distant  from  the  plane  M  N. 

.'.BG  is  II  to  MN,  §  432 

(a  line  is  II  to  a  plane  if  all  its  points  be  equally  distant  from  the  plane). 

Q.  E.  D. 


LINES   AND   PLANES.  261 


Proposition  IX.     Theorem. 

461.    If  two  planes  be  perpendicular  to  the  same  straight 
line  they  are  parallel. 
P 

7 


I 


Q 


In 


Let  the  two  planes  M N  and  PQ  be  perpendicular  to 
the  straight  line  A  B. 

We  are  to  prove  P  Q  II  to  M  N. 

From  any  point  G  in  the  plane  P  Q  draw  G  D  _L  to  M  N. 

Join  B  C. 

BO  is  J_  to  A  B,  §430 

{if  a  straight  line  be  A.  to  a  jylane  it  is  JL  to  every  line  of  the  plane  drawn 
through  its  foot). 

.'.  B  C  is  II  to  the  plane  MN,  §  460 

{if  a.  straight  line  and  a  plane  be  ±  to  the  same  straight  line  they  are  II ). 

.•.CiHsequalto^,  §  432 

{if  a  straight  line  be  II  to  a  plane,  all  its  points  are  equally  distant  from  tlie 

plane). 

Since  G  is  any  point  in  the  plane  P  Q,  all  the  points  in 
the  plane  P  Q  are  at  equal  distances  from  M N. 

.'.PQ  is  II  to  MN,  §  434 

{two  x>lanes  arc  II  if  all  the  points  of  either  be  equally  distant  from  the  other). 

Q.  E.  D. 


262 


GEOMETRY.  —  BOOK  VI. 


Proposition  X.     Theorem. 

462.  If  two  angles  not  in  the  same  plane  have  their 
sides  respectively  parallel  and  lying  in  the  same  direction, 
they  are  equal. 

M 


Let  A  A  and  A'  be  respectively  in  the  planes  M  N  and 
P  Q  and  have  A  D  parallel  to  A'  D'  and  A  C  parallel 
to  A'  C  and  lying  in  the  same  direction. 

We  are  to  prove  Z.  A  =  Z  A'. 

Take  AD  =  A' D1  and  A  C  =  A'  C. 
Join  A  A',  D  D',  C C,  CD,  C D1. 
Since  A  D  is  equal  and  II  to  A'  D',  the  figure  ADD'  A' 
is  a  O,  §  136 

.\AA'  =  DD'.  §134 

In  like  manner  AA'  =  CC', 

.'.  CCf  =  DD'.  Ax.  1 

Also,  since  0  C  and  D  D1  are  respectively  II  to  A  A',  they 

are  II  to  each  other,  §  459 

(two  straight  lines  II  to  a  third  straight  line  not  in  their  own  plane  are  II  to 

each  other). 

.'.CDD'C'iBa.n.  §  136 

.'.CD  =  C'D',  §134 

.\AADC  =  AA'D'C',  §108 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 
.'./.A  =  ZA', 

(being  homologous  A  of  equal  A). 

Q.  E.  D. 

463.  Corollary.  If  two  angles  lie  in  different  planes  and 
have  their  sides  parallel  and  extending  in  the  same  direction,  the 
planes  are  parallel.  For  the  intersecting  lines,  A  C  and  A  D, 
which  determine  the  plane  M N  are  parallel  respectively  to  the 
lines  A'  C  and  A1  D'  which  determine  the  plane  P  Q,  therefore 
the  planes  are  determined  parallel. 


LINES    AND    PLANES. 


263 


Proposition  -XI.     Theorem. 
464.  Two  parallel  lines  comprehended  between  two  par- 
allel planes  are  equal. 


Let  the  two  parallel  lines  AB  and  C D  be  included 
between  the  parallel  planes  M N  and  P  Q. 

We  are  to  prove  A  B  =  C  D. 

IfAB  and  CD  be  J_  to  the  two  II  planes  they  are  equal,  §  434 
{if  two  planes  be  II,  all  the  points  of  either  are  equally  distant  from  tlie  other). 

If  A  B  and  G  Z)  be  not  JL  to  the  two  II  plane3,  draw  from 
the  points  A  and  C  the  lines  A  E  and  C  F  _L  to  the  plane  M  N. 

A  Bis  II  to  CF,  §458 

(two  lilies  ±  to  tlie  same  plane  are  II ). 

Join  BE  and  D F. 

In  AAEB&nd  C F D, 

AE=CF,  §434 

ZAEB  =  ZCFD,  §430 

(if  a  straight  line  be  ±  to  a  plane  it  is  ±  to  any  line  of  the  plane  drawn 
through  its  foot)  ; 

and  ZBAE  =  ZDCF,  §462 

(if  two  A  not  in  the  same  plane  have  their  sides  II  and  lying  in  the  same 
direction  they  are  equal). 

.'.AAEB  =  A  CFD,  §  107 

{having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and  two 
adj.  A  of  the  other). 


Hence  A  B  =  0  B, 

(being  homologous  sides  of  equal  ▲ ). 


Q.  E.  D. 


264 


GEOMETRY.  —  BOOK    VI. 


Proposition  XII.     Theorem. 

465.    The  intersections  of  two  parallel  planes  by  a  third 
plane  are  parallel  lines. 


Let  the  plane  0  S  intersect  the  parallel  planes  P  Q 
and  M N  in  the  lines  AC  and  B  D  respectively. 

We  are  to  prove  AC  II  to  B  D. 

Through  the  points  A  and  C  draw  the  II   lines  A  B  and 
C  D  in  the  plane  0  S. 

Now  AB  =  CD, 

(II  lines  comjjrehended  between  II  planes  are  equal). 


.'.  ABC D  is  aO, 

(having  two  sides  equal  and  II ). 

.'.AC  is  II  to  BD, 
(being  opposite  sides  of  a  O  ). 


§  464 
§  136 
§125 


Q.  E.  D. 


LINES    AND    PLANES. 


265 


Proposition  XIII.     Theorem. 

466.  If  a  straight  line  be  perpendicular  to  one  of  two 
parallel  planes  it  is  perpendicular  to  the  other. 


Let  MN  and  PQ  be  parallel  planes  and  A  B  be  per- 
pendicular to  PQ. 

We  are  to  prove        A  B  _L  to  M  N. 

Let  two  planes  embracing  AB  intersect  the  planes  M  N  and 
P  Q  in  A  C,  B  E  and  A  D,  B  F  respectively. 

Then  A  C  is  II  to  B  E  and  A  D  to  B  F,  §465 

(the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines). 

But  EB  and  FB  are  J_  to  A  B,  §  430 

(if  a  straight  line  be  ±  to  a  plane  it  is  ±  to  every  straight  line  of  the  plane 
drawn  through  its  foot). 

.-.AC  and  A  D  which  are  respectively  II  to  BE  and  B  F 
are  ±.toAB,  §  67 

(if  a  straight  line  be  ±  to  one  of  two  II  lines,  it  is  A.  to  the  other). 

.-.  A  B  is  _L  to  MN,  §  449 

(if  a  line  be  _L  to  two  straight  lines  in  a  plane  drawn  through  its  foot  it  is  _L 
to  the  plane). 

Q.  E.  D. 

467.  Corollary.  If  two  planes  be  parallel  to  a  third  plane 
they  are  parallel  to  each  other.  For,  every  line  perpendicular  to 
this  third  plane  is  perpendicular  to  the  other  planes ;  and  two 
planes  perpendicular  to  a  straight  line  are  parallel. 


266  GEOMETRY. BOOK   VI. 


Proposition  XIV.     Theorem. 

468.  If  a  straight  line  be  parallel  to  another  straight 
line  drawn  in  a  plane,  it  is  parallel  to  the  plane. 


7 


ML ! I i 

<      — t r—^F 

N 
Let  AG  be  parallel  to  the  line  B I)  in  the  plane  M N. 

We  are  to  prove  AG  II  to  the  plane  MN. 

From  A  and  G,  any  two  points  in  A  G,  draw  A  B  and  G  D 
±to£D,  and  A  E  and  G  F  _L  to  the  plane  M  N. 

Join  BE  and  D F. 

Now  ABkHi  to  GD,  §  65 

{two  straight  lines  ±  to  the  same  line  art  II ). 

Also  AB  =  CD,  §135 

(II  lines  comprehended  between  II  lines  are  equal), 

and  A  E  is  \\  to  GF,  §458 

(two  straight  lines  ±  to  the  same  plane  are  II ). 

.-.  Z  BAE  =  Z  £>GF,  §462 

(if  two  A  not  in  the  same  plane  have  tlieir  sides  II  and  lying  in  the  same 
direction,  they  are  equal). 

.-.  rt.  AAEB  =  it.  A  GFD,  §110 

(two  rt.  A  are  equal  when  an  acute  Z  and  the  hypotenuse  of  the  one  are 
equal  respectively  to  an  acute  A  and  the  hypotenuse  of  the  other). 

r.AE=GF, 

(being  homologous  sides  of  equal  A). 

Now  since  the  points  A  and  C,  any  two  points  in  the  line 
A  G,  are  equally  distant  from  the  plane  MN, 

all    the    points   in    A  G  are    equally    distant    from    the 
plane  MN. 

.'.  A  G  is  II  to  the  plane  MN.  §  432 

Q.  E.  D. 


LINES    AND   PLANES. 


267 


Proposition  XV.     Theorem. 
469.   If  two  straight  lines  be  intersected  by  three  par- 
allel planes  their  corresponding  segments  are  proportional. 


^^^     A 

C 

/ 

P  1 

N 

jy*^     e\ 

1 

r\ 

Q 

/                          fj[ 

\ 

\J 

/            B                                                    "/c 

Let  A  B  and  C  D  be  intersected  by  the  parallel  planes 
MN,  PQ,RS,  in  the  points  A,  E,  B,  and  C,  F,  D. 

AE  ^  OF 
EB~  FD' 


We  are  to  prove 


Draw  A  D  cutting  the  plane  P  Q  in  G. 
Join  E  G  and  FG. 

Then  E  G  is  II  to  B  D,  §465 

{the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines). 

•    AE.  -AG_ 
"  EB  ~  GD' 

(a  line  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those 
sides  proportionally). 


§275 


Also, 


GFis  II  to  AC, 

.    CF 
FD  ~ 

A  G 
GD 

.    AE  = 
'  '  EB 

CF 
FD 

§465 
§275 

Ax.  1 

Q.  E.  D. 


268 


GEOMETRY. BOOK    VI. 


On  Dihedral  Angles. 

470.  Def.  The  amount  of  rotation  which  one  of  two  inter- 
secting planes  must  make  about  their  intersection  in  order  to 
coincide  with  the  other  plane  is  called  the  Dihedral  angle  of 
the  planes. 

The  Faces  of  a  dihedral  angle  are  the  intersecting  planes. 

The  Edge  of  a  dihedral  angle  is  the  intersection  of  its  faces. 

The  Plane  angle  of  a  dihedral  angle  is  the  plane  angle 
formed  by  two  straight  lines,  one  in  each  plane,  perpendicular 
to  the  edge  at  the  same  point. 

Thus,  in  the  diagram, 
C-A  B-D  is  a  dihedral  an- 
gle, CB  and  DA  are  its 
faces,  A  B  is  its  edge,  0  PH 
is  its  plane  angle  if  OP 
and  HP  in  the  faces  be 
perpendicular  to  the  edge  A  B  at  the  same  point  P. 

471.  The  plane  angle  of  a  dihedral  angle  has  the  same  mag- 
nitude from  whatever  point  in  the  edge  we  draw  the  perpendicu- 
lars. For  every  pair  of  such  angles  have  their  sides  respectively 
parallel  (§  65),  and  hence  are  equal  (§  462). 

Two  equal  dihedral  angles,  DA  B-C,  and  D-A  B-E',  have 
corresponding  equal  plane  angles,    DAG  and 
DAE.      This   may  be   shown  by  superposi- 
tion. 

Any  two   dihedral   angles,   C-A  B-E'  and 
E-A  B-H',   have  the  same  ratio  as  their  corre- 
sponding plane  angles,  C  A  E  and  E  A  H.    This 
may  be  shown  by  the  method  employed  in  ^ 
§200  and  §201. 

Hence  a  dihedral  angle  is  measured  by  its 
plane  angle. 

It  must  be  observed  that  the  sides  of  the 
plane  angle  which  measures  the  dihedral  angle  must  be  perpendic- 
ular to  the  edge.    Thus  in  the  rectangular  solid  A  H,  Fig.  1,  the 


SOLID    ANGLES. 


269 


dihedral  angle  F-B  A-H,  is  a  right  dihedral  angle,  and  is  meas- 
ured by  the  angle  CED,  if  its  sides  CE  and  ED,  drawn  in 
the  planes  ^Li^and  AG  respectively,  be  perpendicular  to  AB. 
But  angle  C'E'D',  drawn  as  represented  in  the  diagram,  is 
acute,  while  angle  C"EnD",  drawn  as  represented,  is  obtuse. 


F 

V 

D          D" 

\ 

\ 

II 


Fig.  1. 


Fig.  2. 


Many  properties  of  dihedral  angles  can  be  established  which 
are  analogous  to  propositions  relating  to  plane  angles.  Let  the 
student  prove  the  following  : 

1.  If  two  planes  intersect  each  other,  their  vertical  dihedral 
angles  are  equal. 

2.  If  a  plane  intersect  two  parallel  planes,  the  exterior- 
interior  dihedral  angles  are  equal ;  the  alternate-interior  dihedral 
angles  are  equal ;  the  two  interior  dihedral  angles  on  the  same 
side  of  the  secant  plane  are  supplements  of  each  other. 

3.  When  two  planes  are  cut  by  a  third  plane,  if  the  exterior- 
interior  dihedral  angles  be  equal,  or  the  alternate  dihedral  angles 
be  equal,  or  the  two  interior  dihedral  angles  on  the  same  side  of 
the  secant  plane  be  supplements  of  each  other,  and  the  edges 
of  the  dihedrals  thus  formed  be  parallel,  the  two  planes  are 
parallel. 

4.  Two  dihedral  angles  are  equal  if  their  faces  be  respec- 
tively parallel  and  lie  in  the  same  direction,  or  opposite  direc- 
tions, from  the  edges. 

5.  Two  dihedral  angles  are  supplements  of  each  other  if 
two  of  their  faces  be  parallel  and  lie  in  the  same  direction, 
and  the  other  faces  be  parallel  and  lie  in  the  opposite  direc- 
tion, from  the  edges. 


270 


GEOMETRY. BOOK   VI. 


Proposition  XVI.     Theorem. 

472.    If  a  straight  line  be  perpendicular  to  a  plane 
every  plane  embracing  the  line  is  perpendicular  to  that  plane. 


M 


I 


B 


I 


D 
Let  AB  be  perpendicular  to  the  plane  MN 

We  are  to  prove  any  plane}  P  Q,  embracing  A  B,  perpen- 
dicular to  M  N. 

At  B  draw,  in  the  plane  MN,  B  C  J_  to  the  intersection  D  Q. 

Since  A  B  is  J_  to  MN,  it  is  _L  to  D  Q  and  B  G,         §  430 

(if  a  straight  line  be  A.  to  a  plane,  it  is  ±  to  every  straight  line  in  that  plane 
draum  through  its  foot). 

Now  /.ABC  is  the  measure  of  the 

dihedral  Z  P-D  Q-N.  §  470 

But  Z  ABC  is  sl  right  angle, 

.'.  the  Z  P-D  Q-N  is  a  right  dihedral, 

.'.PQis  X  to  MN. 

Q.  E.  D. 


SOLID    ANGLES. 


271 


Proposition  XVII.     Theorem. 

473.  If  two  planes  be  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their 
intersection  is  perpendicular  to  the  other  plane. 


A 


.1/ 


vz         7 

*  c  In 


D 

bet  the  planes  M N  and  PQ  be  perpendicular  to  each 
other,  and  at  any  point  B  of  their  intersection  DQ 
let  BA  be  drawn  in  the  plane  PQ,  perpendicular 
to  DQ. 

We  are  to  prove  A  B  _L  to  the  plane  M  N. 

Draw         B 0  in  the  plane  MN 1.  to  DQ. 

Then  Z.  A  B  G  is  a  right  angle, 

{being  the  plane  Z.  of  the  rt.  dihedral  Z  formed  by  the  two  planes). 

,\ABisJ\-to  the  two  straight  lines  D  Q  and  B  C. 

.'.  A  B  is  _L  to  the  plane  M N,  §  449 

{if  a  straight  line  be  ±  to  two  straight  lines  drawn  through  its  foot  in  a 
plane,  it  is  _L  to  the  plane). 


Q.  E.  D 


272 


GEOMETRY. 


BOOK   VI. 


Proposition  XVIII.     Theorem. 

474.  If  two  planes  be  perpendicular  to  each  other,  a 
straight  line  drawn  through  any  point  of  intersection  per- 
pendicular to  one  of  the  planes  will  lie  in  the  other  plane. 

C 


Fig.  1. 


Fig.  2. 


Let  PQ  {Fig.  l)be  perpendicular  to  the  plane  M N,  C  Q 
their  intersection,  and  B  A  be  drawn  through  any 
point  B  in  C  Q  perpendicular  to  the  plane  M N. 

We  are  to  prove  that  B  A  lies  in  the  plane  P  Q. 

At  the  point  B  draw  B  A'  in  the  plane  P  Q  J_  to  the  inter- 
section C  Q. 

The  line  B  A'  will  be  _L  to  the  plane  M N,         §  472 

(if  two  planes  be  ±  to  each  other,  a  straight  line  drawn  in  one  of  them  ±  to 
their  intersection  is  _L  to  the  other). 


Now  B  A  is  J_  to  the  plane  MN\ 

.'.  B  A  and  B  A'  coincide, 


Hyp. 
§447 


(at  a  given  point  in  a  plane  only  one  ±  can  be  erected  to  that  plane). 
But  B  A'  lies  in  the  plane  P  Q ; 

.'.  B  A,  which  coincides  with  BA/,  lies  in  the  plane  P  Q. 

Q.  E.  D. 

Scholium.  Through  a  line  parallel  or  oblique  to  a  plane,  as 
A  C,  Fig.  2,  only  one  plane  can  be  passed  perpendicular  to  the 
given  plane. 


SOLID    ANGLES. 


273 


Proposition  XIX.     Theorem. 

475.  Jf  two  intersecting  planes  be  each  perpendicular  to 
a  third  plane,  their  intersection  is  also  perpendicular  to  that 
plane. 


I 


Let  the  planes  BD  and  BC  intersecting  in  the  line 
A  B  be  perpendicular  to  the  plane  PQ. 

We  are  to  prove      A  B  J_  to  the  plane  P  Q. 

A  perpendicular  erected  at  B,  a  point  common  to  the  three 
planes,  will  lie  in  the  two  planes  B  C  and  B I),  §  473 

(if  two  planes  be  ±  to  each  other,  a  straight  line  drawn  through  any  point 
of  intersection  ±  to  one  of  the  planes  will  lie  in  the  other  plane). 

And,  since  this  _L  lies  in  both  the  planes,  B  C  and  B  D,  it 
must  coincide  with  their  intersection. 


.'.  A  B  is  J_  to  the  plane  P  Q. 


Q.  E.  D. 


476.  Corollary.  If  a  plane  be  perpendicular  to  each  of 
two  intersecting  planes,  it  is  perpendicular  to  the  intersection  of 
those  planes.  • 


274  GEOMETRY.  —  BOOK   VI. 


Proposition  XX.     Theorem. 

477.  Every  point  in  the  plane  which  bisects  a  dihedral 
angle  is  equally  distant  from  the  faces  of  that  angle. 


A 

Let  plane  A  M  bisect  the  dihedral  angle  formed  by 
the  planes  A  D  and  A  C ;  and  let  P E  and  P F  be 
perpendiculars  drawn  from  any  point  P  in  the 
plane  A  M  to  the  planes  A  C  and  A  D. 

We  are  to  prove  P  E  =  P  F. 

Through  P  E  and  P  F  pass  a  plane  intersecting  the  planes 
A  G  and  A  D  in  0  E  and  0  F. 

Join  P  0. 

Now  the  plane  P  E  F  is  J_  to  each  of  the  planes  A  C  and 
A  D,  §  471 

(if  a  straight  line  be  J.  to  a  plane,  any  plane  embracing  the  line  is  A.  to  that 

plane)  ; 

.*.  the  plane  PE F  is  JL  to  their  intersection  A  0.    §  476 

(If  a  plane  be  _L  to  each  of  two  intersecting  planes,  it  is  ±to  the  intersection 
of  these  planes) . 

.-.Z  POE  =  Z  POF, 

(being  measures  respectively  of  the  equal  dihedral  A  M-OA-C  and  M-OA-D). 
.'.rt.  APOE  =  vt.  A  POF,  §  110 

.'.PE  =  PF, 

(being  homologous  sides  of  equal  A  ). 

Q.  E.  D. 


SOLID    ANGLES. 


275 


Supplementary  Propositions. 
Proposition  XXI.     Theorem. 

478.  The  acute  angle  which  a  straight  line  makes  with 
its  own  projection  on  a  plane  is  the  least  angle  which  it  makes 
with  any  line  of  that  plane. 

A 


Let  B  A  meet  the  plane  M N  at  B,  and  let  B  A'  be  its 
projection  upon  the  plane  M  N,  and  BO  any  other 
line  drawn  through  B  in  the  plane. 

We  are  to  prove     Z  ABA'  <Z  ABC. 

Take  BC=BA'. 

Join  A  C. 

In  the  A  A  B  A'  and  A  B  C, 

AB  =  AB, 

BA'  =  BC, 

but  AA'  <  AC, 

(a  ±  is  the  shortest  distance  from  a  point  to  a  plane). 

.\  Z  ABA'  <Z  ABC,  §  116 

{if  two  sides  of  a  Abe  equal  respectively  to  two  sides  of  another,  but  the  third 

side  of  the  first  A  be  greater  than  the  third  side  of  the  second,  ilien  the 

Z  opposite  the  third  side  of  the  first  A  is  greater  than  tlie  Z  ojjposite  the 

third  side  of  the  second). 

Q.  E.  D. 

Exercise.  —  The  angle  included  by  two  perpendiculars  drawn 
from  any  point  within  a  dihedral  angle  to  its  faces,  is  the  supple- 
ment of  the  dihedral  angle. 


Iden. 
Cons. 
§  448 


276  GEOMETRY. BOOK   VI. 

Proposition  XXII.     Theorem. 

479.   If  two  straight  lines  be  not  in  the  same  plane,  one 
and  only  one  common  perpendicular  to  the  lines  can  be  drawn. 
C  E  D 


M    A 


b\      -fl— £--<? 


7 


^;. ?.: ^../at 

Let  AB  and  C  D  be  two  given  straight  lines  not  in 
the  same  plane. 
We  are  to  prove  one  and  only  one  common  perpendicular  to 
the  two  lines  can  be  drawn. 

Since  A  B  and  C  D  are  not  in  the  same  plane  they  are 
not  II,  §  474 

(two  lis  lie  in  the  same  plane). 

Through  the  line  A  B  pass  the  plane  M N  II  to  CD. 

Since  CD  is  II  to  the  plane  M  N,  all  its  points  are  equally 
distant  from  the  plane  M ' N  \  §  432 

hence  CD',  the  projection  of  the  line  C D  on  the  plane 
MN,  will  be  It  to  CD,  §  76 

and  will  intersect  the  line  A  B  at  some  point  as  O. 

Now  since  C  O  is  the  line  which  projects  the  point  C  upon 
the  plane  MN,  it  is  J_  to  the  plane  MN;  §  435 

hence  C  C  is  _L  to  C  D1  and  A  B,  §  430 

(if  a  line  be  ±  to  a  plane,  it  is  ±  to  every  line  drawn  through  its  foot  in  the 

plane). 

Also,  CO  is  J_  to  CD,  §67 

.*.  C  C  is  the  common  J_  to  the  lines  CD  and  A  B. 

Moreover,  line  C  C  is  the  only  common  _L 

For,  if  another  line  E  B,  drawn  between  A  B  and  CD,  could 

be  J_  to  A  B  and  C  D,  it  would  also  be  J_  to  a  line  B  G  drawn 

II  to  CD  in  the  plane  M N,  §  67 

and  hence  _L  to  the  plane  M  N.  §  449 

But  EH,  drawn  in  the  plane  CD1  II  to  CO,  is  _L  to  the 

plane  MN.  §  457 

Hence  we  should  have  two  Je.  from  the  point  E  to  the  plane 

M N,  which  is  impossible,  §  44/ 

.*.  CO  is  the  only  common  J_  to  the  lines  CD  and  A  B. 

Q.  E.  D 


POLYHEDRAL    ANGLES. 


277 


On  Polyhedral  Angles. 


480.  Def.    A  Polyhedral  angle  is  the  extent  of  opening  of 
three  or  more  planes  meeting  in  a  common  point. 

Thus  the  figure  S-A  EC  BE, 
formed  by  the  planes  A  SB, 
BSC,  CSD,  DSE,  ESA, 
meeting  in  the  common  point 
S,  is  a  polyhedral  angle. 

The  point  S  is  the  vertex  of 
the  angle. 

The  intersections  of  the  planes 
SA,  SB,  etc.,  are  its  edges. 

The  portions  of  the  planes 
bounded  by  the  edges  are  its  faces. 

The  plane  angles  A  SB,  BSC,  etc.,  formed  by  the  edges  are 
its  face  angles. 

481.  Def.  Polyhedral  angles  are  classified  as  trihedral,  quad- 
rahedral,  etc.,  according  to  the  number  of  the  faces. 

482.  Def.  Trihedral  angles  are  rectangular,  bi-rectangular,  or 
tri-rectangular,  according  as  they  have  one,  two,  or  three  right 
dihedral  angles. 

483.  Def.  Trihedral  angles  are  scalene,  isosceles,  or  equilateral, 
according  as  the  face  angles  are  all  unequal,  two  equal,  or  three 
equal. 

484.  Def.  A  polyhedral  angle  is  convex,  if  the  polygon  formed 
by  the  intersections  of  a  plane  with  all  its  faces  be  a  convex 
polygon. 

485.  Def.  Two  polyhedral  angles  are  equal  when  they  can 
be  applied  to  each  other  so  as  to  coincide  in  all  their  parts. 

Since  two  equal  polyhedral  angles  coincide  however  far  their 
edges  and  faces  be  produced,  the  magnitude  of  a  polyhedral  angle 
does  not  depend  upon  the  extent  of  its  faces.  But,  in  order  to 
represent  the  angle  in  a  diagram,  it  is  usual  to  pass  a  plane,  as 


278  GEOMETRY.  —  BOOK   VI. 

ABODE,  cutting  all  its  faces  in  the  straight  lines,  A  B,  B  C, 
etc. ;  and  by  the  face  A  S  B  is  meant  the  indefinite  surface  in- 
cluded between  the  lines  S A  and  SB  indefinitely  produced. 

486.  Def.  Two  polyhedral  angles  are  symmetrical  if  they 
have  the  same  number  of  faces,  and  the  successive  dihedral  and 
face  angles  respectively  equal  but  arranged  in  reverse  order. 

Thus,  if  the  edges   AS,   B S, 

a       Ft1 

etc.,    of    the     polyhedral     angle,  * :-N 

S-A  BCD,  be  produced,  there  is  m.S.'jJF 

formed  another  polyhedral  angle,  /  //  y*'* 

S-A1  B1  C  D>,  which  is  symmetri-  l/y'y 

cal  with  the  first,   the  vertex  S  Sj£' 

being  the  centre  of  symmetry.  //n  V^N 

If  we   take    S  A' =  S  A,    and         ////    \  V\ 
through  the  points  A  and  A'  the    A<^  y  (nj       \  d>\^—^.a 

parallel     planes     A  B  C  D    and        \z___V  v. \/ 

A'B'C  D   be   passed,    we   shall        B         C  c  b 

have  SB'  =  SB,  SC'  =  SC,  etc.  For  if  we  conceive  a  third 
parallel  plane  to  pass  through  S,  then  A  A',  B  B',  etc.,  are 
divided  proportionally,  §  469.  And  if  any  one  of  them  be 
bisected  at  S,  the  others  are  also  bisected  at  S.  Hence,  the 
points  A',  B',  etc.,  are  symmetrical  with  A,  B,  etc. 

Moreover,  the  two  symmetrical  polyhedral  angles  are  equal  in 
all  their  parts.  Tor  their  face  angles  A  SB  and  A'  SB',  B  SO 
and  B'  S  C  are  equal  each  to  each,  being  vertical  plane  angles. 
And  the  dihedral  angles  formed  at  the  edges  S  A  and  SA',  SB 
and  SB',  are  equal  each  to  each,  being  vertical  dihedral  angles. 

Now  if  the  polyhedral  angle  S-A'  B'  C  D1  be  revolved  about 
the  vertex  S  until  the  polygon  A'  B'  C  D  is  brought  into  the 
position  abed,  in  the  same  plane  with  ABC D,  it  will  be 
evident  that  while  the  parts  A  SB,  B SC,  etc.,  succeed  each 
other  in  the  order  from  left  to  right,  the  corresponding  equal 
parts  a  Sb,  b  Sc,  etc.,  succeed  each  other  in  the  order  from  rigid 
to  left.  Hence  the  two  figures  cannot  be  made  to  coincide  by 
superposition,  but  are  said  to  be  equal  by  symmetry. 


SOLID   ANGLES.  279 


Proposition  XXIII.     Theorem. 
487.    The  sum   of  any  two  face   angles  of  a  trihedral 
angle  is  greater  than  the  third.       ^ 


Let  S-ABC  be  a  trihedral  angle  in  which  the  face 
angle  ASC  is  greater  than  either  angle  A  S B  or 
angle  BSC. 

We  are  to  prove  Z  ASB  +  Z  BSC  >  Z  ASC. 
In  the  face  A  SG  draw  S D,  making  Z  A  SD  =  Z  A  SB. 
Through  any  point  D  of  S  D  draw  any  straight  line  ADC 
cutting  A  S  and  S  C. 

TakeSB  =  SD. 
Pass  a  plane  through  A  C  and  the  point  B. 
In  the  A  A  S  D  and  A  S  B 

AS=AS,  Iden. 

SD  =  SB,  Cons. 

ZASD  =  ZASB.  Cons. 

.'.AASD  =  A  A  SB,  §  10G 

.'.AD  =  AB, 

{being  homologous  sides  of  equal  A). 

lathe  A  ABC,       AB  +  BOAC. 

Subtract  the  equals  A  B  and  A  D. 
Then  BO  DC. 

Now  in  the  A  BSC  and  DSC 

SB=SD,  Cons. 

SC  =  SC,  Iden. 

but  BO  DC, 

.\Z  BSOZ  DSC.  §116 

.'.ZASB  +  Z  BSC  >  ZASD  +  Z  DSC, 
that  is        ZASB+ZBSOZASC. 


280 


GEOMETRY. 


BOOK  VI. 


Proposition  XXIV.     Theorem. 

488.    The  sum  of  the  face  angles  of  any  convex  polyhe- 
dral angle  is  less  than  four  right  angles. 

S 


Let  the  polyhedral  angle  S  be  cut  by  a  plane,  mak- 
ing the  section  ABC  DE  a  convex  polygon. 

We  are  to  prove  Z  A  SB  +  Z  BSC  etc.  <  4  rt.  A. 

From  any  point  0  within  the  polygon  draw  0  A,0  B,  0  C, 
OD,  OE. 

The  number  of  the  A  having  their  common  vertex  at  0 
will  be  the  same  as  the  number  having  their  common  vertex  at  S. 

.*.  the  sum  of  all  the  A  of  the  A  having  the  common  vertex 
at  S  is  equal  to  the  sum  of  all  the  A  of  the  A  having  the  com' 
mon  vertex  at  0. 

But  in  the  trihedral  A  formed  at  A,  B,  C,  etc. 

ZSAE+  ZSAB>Z  OAE+  Z  0  A  B,      §  487 
(the  sum  of  any  two  face  A  of  a  triliedral  Z  is  greater  than  the  third). 

and  Z  SBA  +  Z  SBOZ  OBA  +  Z  OBC.       §487 

.'.  the  sum  of  the  A  at  the  bases  of  the  A  whose  common 
vertex  is  S  is  greater  than  the  sum  of  the  A  at  the  bases  of  the 
A  whose  common  vertex  is  0. 

.'.  the  sum  of  the  A  at  S  is  less  than  the  sum  of  the  A  at  0. 

But  the  sum  of  the  A  at  0  =  4  rt.  A .  §  34 

.*.  the  sum  of  the  A  at  8  is  less  than  4  rt.  A . 

Q.  E.  D, 


SOLID   ANGLES. 


281 


Proposition  XXV.     Theorem. 
489.   An  isosceles  trihedral  angle  and  its  symmetrical 
trihedral  angle  are  equal. 


Let  S-A  B  C  be  an  isosceles  trihedral  angle,  having; 
ZASB  =  ZBSC.  And  let  S-A'  B'  C  be  its  sym- 
metrical trihedral  angle. 

We  are  to  prove  trihedral ZS-ABC  =  trihedral  Z  S-A' B'  C. 

Revolve  Z  S-A' B' C  about  S  until  SB'  falls  on  SB  and 
the  plane  SB' A'  falls  on  the  plane  SBC. 

Now  the  dihedral  Z  C-SB-A  =  dihedral  Z  A'-SB'-C, 
{being  vertical  dihedral  A ). 

.*.  the  plane  SB'  C  will  fall  on  the  plane  SB  A. 

Now  ZBSC  =  ZBSA,  Hyp. 

and  Z  B'SA'  =  ZBSAf 

(being  vertical  A ). 

.\Z  BSO  =  Z  B'SA';  Ax.  1 

.'.SA'  will  fall  on  SO. 

In  like  manner  S  C  will  fall  on  S  A, 

.'.  the  two  trihedral  A  will  coincide  and  be  equal. 

q.  e.  o. 


282 


GEOMETRY. BOOK    VI. 


Proposition  XXVI.     Theorem. 
490.    Two  symmetrical  trihedral  angles  are  equivalent. 


Let  the  trihedral  Z  S-ABC  and  Z  S-A' B' C    be  sym- 
metrical. 

We  are  to  prove  trihedrat  Z  S-ABC  =c=*  trihedral  Z  S-A'B'C. 

Draw  D' D  making  the  A  DS A,  DSC,  and  DSB  equal. 

Then      ZD'SA'  =  ZD'SC'  =  A  D'SB', 

{being  vertical  A  of  (lie  equal  A  D  S  A,  DSC,  and  D  SB). 

Then  the  trihedral  Z  S-D  CB  =  trihedral  Z  S-D'  C'B'}    §  489 
(tvjo  isosceles  symmetrical  trihedral  A  are  equal). 

And  trihedral  Z  S-D  C  A  =  trihedral  Z  S-D'  C  A', 

and  trihedral  ZS-ADB  =  trihedral  Z  S-A1  D' B'. 

Adding  the  first  two  equalities,  the  polyhedral  Z  S-A  BCD 

ro=  polyhedral  Z  S-A' B' CD1. 

Take  away  from  each  of  these  equals  the  equal  trihedral 
A  S-ADB  and  S-A'  D'  B'. 

Then  trihedral  Z  S-ABC '^  trihedral  Z  S-A' B' C. 

Q.  E.  D. 

491.  Scholium.  If  D  D'  fall  within  the  given  trihedral 
angles  these  trihedral  angles  would  be  composed  of  three  isosceles 
trihedral  angles  which  would  be  respectively  equal,  and  hence 
the  given  trihedral  angles  would  be  equivalent. 

*  The  symbol  (o)  is  to  be  read  "  equivalent  to." 


SOLID   ANGLES.  283 


Exercises. 

1.  If  a  plane  be  passed  through  one  of  the  diagonals  of  a 
parallelogram,  the  perpendiculars  to  the  plane  from  the  extremi- 
ties of  the  other  diagonal  are  equal. 

2.  If  each  of  the  projections  of  a  line  A  B  upon  two  inter- 
secting planes  be  a  straight  line,  the  line  A  B  is  a  straight  line. 

3.  The  height  of  a  room  is  eight  feet,  how  can  a  point  in 
the  floor  directly  under  a  certain  point  in  the  ceiling  be  deter- 
mined with  a  ten-foot  pole  ] 

4.  If  a  line  be  drawn  at  an  inclination  of  45°  to  a  plane,  what 
is  the  greatest  angle  which  any  line  of  the  plane,  drawn  through 
the  point  in  which  the  inclined  line  pierces  the  plane,  makes 
with  the  line. 

5.  Through  a  given  point  pass  a  plane  parallel  to  a  given 
plane. 

6.  Find  the  locus  of  points  in  space  which  are  equally  distant 
from  two  given  points. 

7.  Show  that  the  three  planes  embracing  the  edges  of  a  tri- 
hedral angle  and  the  bisectors  of  the  opposite  face  angles  re- 
spectively intersect  in  the  same  straight  line. 

8.  Find  the  locus  of  the  points  which  are  equally  distant  from 
the  three  edges  of  a  trihedral  angle. 

9.  Cut  a  given  quadrahedral  angle  by  a  plane  so  that  the 
section  shall  be  a  parallelogram. 

10.  Determine  a  point  in  a  given  plane  such  that  the  sum  of 
its  distances  from  two  given  points  on  the  same  side  of  the  plane 
shall  be  a  minimum. 

11.  Determine  a  point  in  a  given  plane  such  that  the  differ- 
ence of  its  distances  from  two  given  points  on  opposite  sides  of 
a  plane  shall  be  a  maximum. 


284 


GEOMETRY. 


BOOK   VI. 


Proposition  XXVII.     Theorem. 

492.  Two  trihedral  angles  are  equal  or  symmetrical 
when  the  three  face  angles  of  the  one  are  respectively  equal  to 
the  three  face  angles  of  the  other. 

S'  S  S> 


In  the  trihedral  A  S  and  S',  let  Z  A  S  B  =  Z  A'  S'  B>, 
Z  ASC  =  ZA'S'C')  and  Z  BSC  =  ZB'S'C. 

We  are  to  prove  that  the  homologous  dihedral  angles  are  equal, 
and  hence  the  trihedral  angles  S  and  S'  are  either  equal  or 
symmetrical. 

On  the  edges  of  these  angles  take  the  six  equal  distances 
SA,SB,  SC,S'A',S'B',S'C'. 

Draw  A  B,  B  C,  A  C,  A'B',  B'C,  A'C. 

The  homologous  isosceles  A  SAB,  S>  A' B',  SAC,  S1  A'  C, 
SBC,  S'B' C  are  equal,  respectively.  §  106 

.'.AB,AC,BC  equal  respectively  A'B',  A'  C,  B'  C, 

{being  homologous  sides  of  equal  A). 

.-.  A  ABC  =  A  A'B' O.  §  108 

At  any  point  D  in  SA  draw  D  E  and  D  F  _L  to  SA  in  the 
faces  AS  B  and  ASC  respectively. 

These  lines  meet  A  B  and  A  C  respectively, 
(since  the  A  SAB  and  SAC  are  acute,  each  being  one  of  the  equal  A  of  an 
isosceles  A). 

Join  EF. 

QnS'A'takQA'D'  =  AD. 


SOLID   ANGLES.  285 


Draw  D'  E'  and  D'  F  in  the  faces  A1 S'  B'  and  A' S'  C  re- 
spectively ±  to  S'  A',  and  join  E'  F'. 
In  the  rt.  A  A  DE  and  A'  D'  E' 

AD  =  A>D',  Cons. 

ZDAE=Z  D'A'E', 
(being  homologojts  A  of  the  equal  &  SAB  and  &  A*  B1). 

.-.rt.  A  ADE  =  vt.  A  A'  D'  E\  §  111 

.-.  AE  =  A'E'  and  D  E  =  D'  E't 

{being  homologous  sides  of  equal  &). 

In  like  manner  we  may  prove  AF '  =  A' F'  and  DF~D'F. 

Hence  in  the  A  A  E  F  and  A1  E'  F'  we  have 

A  E  and  A  F  =  respectively  A'  E'  and  A1  F, 

and  ZEAF=ZE,A/F, 

(being  homologous  A  of  the  equal  A  ABC  and  Af  B1  C). 

.:AAEF  =  AA'E'F',  §106 

.-.  EF=  E'F 
(being  homologous  sides  of  the  equal  A  AEF  and  A1  E'  F). 

Hence,  in  the  A  E  D  F  and  E'  D*  F  we  have 

ED,DF,  and  EF=  respectively  E' D',  b' P,'slti&  E1  F'. 

.\AEDF=A  E'jyP,  §  108 

.-.  Z  EDF  =  Z  E'D'P, 

(being  homologous  A  of  equal  &). 

.-.  the  dihedral  Z  B-A  S-C  =  dihedral  Z  B'-A'  S'-C, 
(since  A  E  D  F  and  E'  D'  F,  the  measures  of  these  dihedral  A,  are  equal). 

In  like  manner  it  may  be  proved  that  the  dihedral 
A  A-B  S-C  and  A-C  SB  are  equal  respectively  to  the  dihedral 
A  A'-B'S'-C  and  A'-C  S'-B'. 

Q.  E.  D. 

This  demonstration  applies  to  either  of  the  two  figures  de- 
noted by  S'-A'  B'  C,  whicli  are  symmetrical  with  respect  to  each 
other.  If  the  first  of  these  figures  be  given,  S  and  S'  are  equal, 
for  they  can  be  applied  to  each  other  so  as  to  coincide  in  all  their 
parts.     If  the  second  be  given,  S  and  S'  are  symmetrical.    §  486 


BOOK   VII. 

POLYHEDRONS,   CYLINDERS,  AND  CONES. 


General  Definitions. 


493.  Def.  A  Polyhedron  is  a  solid  bounded  by  four  or 
more  polygons. 

A  polyhedron  bounded  by  four  polygons  is  called  a  tetra- 
hedron; by  six,  a  hexahedron;  by  eight,  an  octahedron;  by  twelve, 
a  dodecahedron;  by  twenty,  an  icosahedron. 

494.  Def.  The  Faces  of  a  polyhedron  are  the  bounding 
polygons. 

495.  Def.  The  Edges  of  a  polyhedron  are  the  intersec- 
tions of  its  faces. 

496.  Def.  The  Vertices  of  a  polyhedron  are  the  intersec- 
tions of  its  edges. 

497.  Def.  A  Diagonal  of  a  polyhedron  is  a  straight  line 
joining  any  two  vertices  not  in  the  same  face. 

498.  Def.  A  Section  of  a  polyhedron  is  a  polygon  formed 
by  the  intersection  of  a  plane  with  three  or  more  faces. 

499.  Def.  A  Convex  polyhedron  is  a  polyhedron  every 
section  of  wnich  is  a  convex  polygon. 

500.  Def.  The  Volume  of  a  polyhedron  is  the  numerical 
measure  of  its  magnitude  referred  to  some  other  polyhedron  as  a 
unit  of  measure. 

501.  Def.  The  polyhedron  adopted  as  the  unit  of  measure 
is  called  the  Unit  of  Volume. 

502.  Def.  Similar  polyhedrons  are  polyhedrons  which 
have  the  same  form. 

503.  Def.  Equivalent  polyhedrons  are  polyhedrons  which 
have  the  same  volume. 

504.  Def.  Equal  polyhedrons  are  polyhedrons  which  have 
the  same/orra  and  volume. 

On  Prisms. 

505.  Def.  A  Prism  is  a  polyhedron  two  of  whose  faces 
are  equal  and  parallel  polygons,  and  the  other  faces  are  parallelo- 
grams. 


PRISMS. 


287 


506.  Def.    The  Bases  of  a  prism 
are  the  equal  and  parallel  polygons. 

507.  Def.    The  Lateral  faces  of 
a  prism  are  all  the  faces  except  the 


508.  Def.  The  Lateral  or  Con- 
vex Surface  of  a  prism  is  the  sum  of 
its  lateral  faces. 

509.  Def.  The  Lateral  edges 
of  a  prism  are  the  intersections  of  its 
lateral  faces ;  the  Basal  edges  of  a 
prism  are  the  intersections  of  the  bases 
with  the  lateral  faces. 

510.  Def.  Prisms  are  triangular,  quadrangular,  pentag- 
onal, etc.,  according  as  their  bases  are  triangles,  quadrangles, 
pentagons,  etc. 

511.  Def.  A  Right  prism  is  a  prism  whose  lateral  edges 
are  perpendicular  to  its  bases. 

512.  Def.  An  Oblique  prism  is  a  prism 
whose  lateral  edges  are  oblique  to  its  bases. 

513.  Def.  A  Regular  prism  is  a  right 
prism  whose  bases  are  regular  polygons,  and 
hence  its  lateral  faces  are  equal  rectangles. 

514.  Def.  The  Altitude  of  a  prism  is 
the  perpendicular  distance  between  the  planes 
of  its  bases.  The  altitude  of  a  right  prism  is 
equal  to  any  one  of  its  lateral  edges. 

515.  Def.     A  Truncated  prism  is  a  por- 
tion of  a  prism  included  between  either  base 
and  a  section  inclined  to  the  base  and  cutting  R|QHT  pr|sm. 
all  the  lateral  edges. 

516.  Def.  A  Right  section  of  a  prism  is  a  section  perpen- 
dicular to  its  lateral  edges. 

517.  Def.  A  Parallelopiped  is  a  prism  whose  bases  are 
parallelograms. 

518.  Def.  A  Right  parallelopiped  is  a  parallelopiped  whose 
lateral  edges  are  perpendicular  to  its  bases  ;  hence  its  lateral  faces 
are  rectangles. 

519.  Def.  An  Oblique  parallelopiped  is  a  parallelopiped 
whose  lateral  edges  are  oblique  to  its  bases. 

520.  Def.  A  Rectangular  parallelopiped  is  a  right  paral- 
lelopiped whose  bases  are  rectangles. 

521.  Def.  A  Cube  is  a  rectangular  parallelopiped  all  of 
whose  faces  are  squares. 


288  GEOMETRY.  —  BOOK   VII. 

Proposition  I.     Theorem. 

522.    The  sections  of  a  prism  made  by  parallel  plat 
are  equal  polygons. 


Let   the  prism  A  D  be  intersected   by  the  parallel 
planes  G  K,  G'  K'. 

We  are  to  prove  section  G  H I K  L  =  section  G' H' I'  K' L'. 

G  H,  III,  IK,  etc.,  are  parallel  respectively  to  G'  H',  H'  I', 
I'K',etc,  §465 

{the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines). 

.'.  AG  HI,  H IK,  etc.,  are  equal  respectively  to  A  G1  H'  I', 
H,rKr9  etc.,  §462 

{two  A  not  in  the  same  plane,  having  their  sides  respectively  parallel  and 
lying  in  the  same  direction,  are  equal). 

Also,  sides  GH,  HI,  IK,  etc.,  are  equal  respectively  to 
G'W,  11'1',1'K1,  etc.,  §  135 

( II  lines  comprehended  between  II  lines  are  equal). 

.'.  section  GHIKL  =  section  G'  W  I>  K'  L',         §  155 

(being  mutually  equiangular  and  equilateral). 

Q.  E.   D. 

523.  Corollary.  Any  section  of  a  prism  parallel  to  the 
base  is  equal  to  the  base ;  and  all  right  sections  of  a  prism  are 
equal. 


prisms.  289 


Proposition  II.     Theorem. 
524.    The  lateral  area  of  a  prism  is  equal  to  the  product 
of  a  lateral  edge  by  the  perimeter  of  the  right  section. 


E> 


D> 


Let  GH I KL  be  a  right  section  of  the  prism  AD'. 

We  are  to  prove  lateral  area  of  prism  A  D1  =  A  A'  X  perim- 
eter G H I K L. 

Consider  the  lateral  edges  A  A',  B  B',  etc.,  to  be  the  bases 
of  the  U]  AB',  B  C'}  etc.,  which  form  the  convex  surface  of  the 
prism. 

Then  the  altitudes  of  these  HI  will  be  the  J»  GH,  HI, 
IK,  etc., 

and  the  area  of  each  O  is  the  product  of  its  base  and  alti- 
tude. §  321 
Now  the  bases  of  these  Z17  are  all  equal,  §  464 
(II  lines  comprehended  between  II  planes  are  equal)  ; 
and  the  sum  of  the  altitudes  GH,  HI,  IK,  etc.,  is  the  perimeter 
of  the  right  section. 

Hence,  the  sum  of  the  areas  of  these  ZI7  is  the  product  of  a 
lateral  edge  A  A'  by  the  perimeter  of  the  right  section. 

That  is,  the  lateral  area  of  the  prism  is  equal  to  the  product 
of  a  lateral  edge  by  the  perimeter  of  a  right  section. 

Q.  E.  D. 

525.  Corollary.  The  lateral  area  of  a  right  prism  is  equal 
to  the  altitude  multiplied  by  the  perimeter  of  the  base. 


290  GEOMETRY.  —  BOOK   VII. 

Proposition  III.     Theorem. 
526.    Two  prisms  are  equal  if  the  three  faces  including 
a  trihedral  angle  of  the  one  be  respectively  equal  to  the  three 
corresponding  faces  including  a  trihedral  angle  of  the  other , 
and  similarly  placed. 

J  J* 


F 


Let  AD,  AG,  A  J,  be  respectively  equal  to  A'  D', 

A '  J1,  an  d  similarly  pla  ced. 

We  are  to  prove  prism  A  1  =  prism  A'  I'. 

Now  trihedral  Z  A  =  trihedral  Z  A1,  §  492 

(two  trihedrals  arc  equal,  when  the  three  face  A  of  the  one  are  equal  respec- 
tively to  tlic  three  face  A  of  the  other  and  are  similarly  placed). 

Apply  trihedral  Z  A  to  trihedral  Z  A'. 

Then  the  base  A  D  will  coincide  with  the  base  A'  D', 

face  A  G  with  A'  G', 

and  face  A  J  with  A'  J' ; 

.'.  FG  will  coincide  with  F'G',  and  F  J  with  FJ'. 

.'.  the  upper  bases,  F I  and  F'  I1,  will  coincide, 
(being  equal  polygons,  since  they  arc  equal  to  the  equal  lower  bases). 

.*.  the  remaining  edges  will  coincide, 

(their  extremities  being  the  same  points). 

.'.  the  prisms  will  coincide  and  be  equal. 

Q.  E.  D. 

527.  Corollary  1.  Two  truncated  prisms  are  equal,  if 
the  three  faces  including  a  trihedral  of  the  one  be  respectively 
equal  to  the  three  faces  including  a  trihedral  of  the  other,  and 
be  similarly  placed. 

528.  Cor.  2.  Two  right  prisms  having  equal  bases  and 
altitudes  are  equal.  If  the  faces  be  not  similarly  placed,  if  one 
be  inverted,  the  faces  will  be  similarly  placed  and  the  prisms  can 
be  made  to  coincide. 


PRISMS.  291 


Proposition  IV.     Theorem. 

529.  An  oblique  prism  is  equivalent  to  a  right  prism 
whose  bases  are  equal  to  right  sections  of  the  oblique  prism, 
and  whose  altitude  is  equal  to  a  lateral  edge  of  the  oblique 
prism. 

B/' 


B     -        C 
Let  A  D'  be  an  oblique  prism,  and  F I  a  right  section. 

Complete  the  right  prism  F  I',  making  its  edges  equal  to 
those  of  the  oblique  prism. 

We  are  to  prove  oblique  prism  A  D'  ^=  right  prism  F  I'. 

In  the  solids  A  I  and  A'  I' 

trihedral  Z  A  =  trihedral  Z  A',  §  492 

{two  trihedrals  are  equal  when  three  face  A  of  the  one  are  respectively  equal 
to  three  face  A  of  the  other,  and  are  similarly  placed). 

Xow  face  A  D  =  face  A1  D',  §  505 

(being  the  two  bases  of  the  oblique  prism  A  D')  ; 

face  A  J  =  face  A'  J',  Cons. 

and  face  A  G  =  face  A'  G'.  Cons. 

.'.  solid  AI=  solid  AT,  §  527 

(two  truncated  prisms  are  equal  wlicn  the  three  faces  including  a  trihedral 
of  the  one  are  respectively  equal  to  tJie  three  faces  including  a  trihedral 
of  the  other,  and  are  similarly  placed). 

To  each  of  these  equal  solids  add  the  solid  F D'. 
Then  oblique  prism  A  D'  *>  right  prism  F  I'. 

Q.  E.  D. 


292 


GEOMETRY. BOOK    VII. 


Proposition  V.     Theorem. 

530.    Any  two  opposite  faces  of  a  parallelopiped  are 
equal  and  parallel. 


Let  AG  be  a  parallelopiped. 
We  are  to  prove  faces  A  F  and  D  G  equal  and  parallel. 

Since  A  GYisaO,  §  517 

A  B  and  D  C  are  equal  and  II  line9.  §  125 

Also,  since  A  H  is  a  O,  §  505 

A  E  and  D  H  are  equal  and  II  lines.  §  125 

.'.Z  EAB  =  Z  HDC,  §462 

(two  A  not  in  the  same  plane  having  their  sides  II  and  lying  in  the  same 
direction  are  equal). 


.'.  face  AF=  face  D  G. 


§  140 
§  463 


allel. 


Moreover,  face  A  F  is  II  to  D  G 

the  same  plane  have  their  sides  II  an 
direction  their  planes  are  parallel). 

In  like  manner  we  may  prove  A  H  and  B  G  equal  and  par- 


(if  two  A  not  in  the  same  plane  have  their  sides  II  and  lying  in  the  same 
direction  their  planes  are  parallel,). 


Q.  E.  D. 


531.  Scholium.  Any  two  opposite  faces  of  a  parallelo- 
piped may  be  taken  for  bases,  since  they  are  equal  and  parallel 
parallelograms. 


prisms.  293 


Proposition  VI.     Theorem. 

532.  The  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelopiped  divides  the  parallelopiped  into  two 
equivalent  triangular  prisms. 

II 


Let  the  plane  A  E  GC  pass  through  the  opposite  edges 
A  E  and  C  G  of  the  parallelopiped  A  G. 

We  are  to  prove  that  the  parallelopiped  A  G  is  divided  into 
two  equivalent  triangular  prisms,  A  B  C-F,  and  A  D  C-H. 

Let  I J KL  be  a  right  section  of  the  parallelopiped  made 
by  a  plane  ±  to  the  edge  A  E. 

The  intersection  IK  of  this  plane  with  the  plane  A  EGC 
is  the  diagonal  of  the  O  I J KL. 

.-.A  IKJ=A  IKL.  §  133 

But  prism  A  B  C-F  is  equivalent  to  a  right  prism  whose 
base  is  UK  and  whose  altitude  is  A  E,  §  529 

(any  oblique  prism  is  ^  to  a  right  prism  whose,  bases  arc  equal  to  right  sec- 
tions of  the  oblique  prism,  and  whose  altitude  is  equal  to  a  lateral  edge 
oftlic  oblique  prism). 

The  prism  A  D  C-H  is  equivalent  to  a  right  prism  whose 
base  is  ILK,  and  whose  altitude  is  A  E.  §  529 

Now  the  two  right  prisms  are  equal,  §  528 

{two  right  prisms  having  equal  bases  and  altitudes  arc  equal). 

.'.ABC-Fo  ADC-H. 

Q.  E.  D 


294 


GEOMETRY. BOOK  VII. 


Proposition  VII.     Theorem. 

533.    Two  rectangular  parallelepipeds  having  equal  bases 
are  to  each  other  as  their  altitudes. 


/ 

B 
/              P              J 

i 

<                               / 

& 

/        pi       J 

m    ( 

'1  , 

/                                / 

/ 

'                               / 

/ 

/                                / 

)* 

/A                            ) 

1 


Let  AB  and  A'B'  be  the  altitudes  of  the  two  rectangu- 
lar parallelopipeds,  P,  and  Pf,  having  equal  bases. 

w         *  p       AB 

We  are  to  prove  —  = . 

P'      A'B' 
CASE  I.  —  When  A  B  and  A1  B'  are  commensurable. 

Find  a  common  measure  m,  of  A  B  and  A'  B'. 
Suppose  m  to  be  contained  in  A  B  5  times,  and  in  A'  B' 
3  times. 

AB      5 
A'B'~3' 

At  the  several  points  of  division  on  AB  and  A' B'  pass 
planes  _L  to  these  lines. 

The  parallelopiped  P  will  be  divided  into  5, 

and  P'  into  3,  parallelopipeds  equal,  each  to  each,    §  528 
(two  right  prisms  having  equal  bases  and  altitudes  are  equal). 

Then  LLt 

P'      3 

.  P  _AB 

"P'~A'Br 


Then  we  have 


PRISMS. 


295 


IV 


Case  II.  —  When  A  B  and  A'  B'  are  incommensurable. 


»£ 


t 


\ 


\ 


Let  A  B  be  divided  into  any  number  of  equal  parts, 

and  let  one  of  these  parts  be  applied  to  A'  B'  as  many  times 
as  A'  B'  will  contain  it. 

Since  A  B  and  A'  B'  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  A'  to  a  point  Df  leaving  a  ra 
mainder  D  B'  less  than  one  of  these  parts. 

Through  D  pass  a  plane  _L  to  A'  B',  and  denote  the  parallel- 
opiped  whose  base  is  the  same  as  that  of  P'y  and  whose  altitude 
is  4' 2)  by  Q. 

Now,  since  A  B  and  A'  D  are  commensurable, 

Q  :P  =  A'D  :AB.  (Case  I.) 

Suppose  the  number  of  parts  into  which  A  B  is  divided  to 
be  continually  increased,  the  length  of  each  part  will  become  less 
and  less,  and  the  point  D  will  approach  nearer  and  nearer  to  B'. 

The  limit  of  Q  will  be  P', 

and  the  limit  of  A'  D  will  be  A'  B'y 

.'.  the  limit  of  Q  :  P  will  be  P'  :  P, 

and  the  limit  of  A'  D  :  A  B  will  be  A'  B'  :  A  B, 

Moreover  the  corresponding  values  of  the  two  variables  Q  :  P 
and  A1  D  :  A  B  are  always  equal,  however  near  these  variables 
approach  their  limits. 

.*.  their  limits  P'  :  P  =  A' B'  :  A  B.  §  199 

Q.  E.  D. 

534.  Scholium.  The  three  edges  of  a  rectangular  parallelo- 
piped  which  meet  at  a  common  vertex  are  its  dimensions.  Hence 
two  rectangular  parallelopipeds  which  have  two  dimensions  in 
common  are  to  each  other  as  their  third  dimensions. 


296 


GEOMETRY. BOOK   VII. 


Proposition  VIII.     Theorem. 

535.    Two  rectangular  parallelopipeds  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


(  Q  ( 

*  I 

1     p>   1 

1 

/p( 

c 

/  I 

(. 

<• 

) 

1    /„  ( 

7 


Let  a,  b,  and c,  and  a',  b',  c,  be  the  three  dimensions  re- 
spectively of  the  two  rectangular  parallelopipeds 
P  and  P. 

We  are  to  prove  —  = . 

*  P'       a'X  b' 

Let  Q  be  a  third  rectangular  parallelopiped  whose  dimen- 
sions are  a',  b  and  c. 

Now  Q  has  the  two  dimensions  b  and  c  in  common  with  P, 
and  the  two  dimensions  a'  and  c  in  common  with  P. 

Then  -=-,  §534 

Q       a''  * 

(two  rectangular  parallelopipeds  which  have  two  dimensions  in  common  are 
to  each  other  as  their  third  dimensions)  ; 

and  |ll. 

P'      ¥ 

Multiply  these  two  equalities  together ; 

P  _     aXb 

P'  ~  a' XV  ' 


then 


§534 


Q.  E.  D. 


536.  Scholium.  This  proposition  may  be  stated  thus  :  two 
rectangular  parallelopipeds  which  have  one  dimension  in  common 
are  to  each  other  as  the  products  of  the  other  two  dimensions. 


PRISMS. 


297 


Proposition  IX.     Theorem. 

537.    Any  two  rectangular  parallelopipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


X 


\    p.   \ 

c 

\   \ 

C 

\            \ 

III 


Let  a,  b,c,  and  a,'  b',  cf,  be  the  three  dimensions  respec- 
tively of  the  two  rectangular  parallelopipeds  P 
and  P. 


We  are  to  prove  =  ___ 


Let  Q  be  a  third  rectangular  parallelopiped  whose  dimen- 
sions are  a,  b,  and  d. 

Then  £  =  £.,  §534 

Q      cf 

{two  rectangular  parallelopipeds  which  have  two  dimensions  in  common  are 

to  each  other  as  their  third  dimensions)  ; 


and 


§536 


Q_  _     aXb 
P'  ~~  a' XV' 

{two  rectangular  parallelopipeds  which  have  one  dimension  in  common  are  to 
each  other  as  tlic  products  of  tlicir  oilier  two  dimensions). 


Multiply  these  equalities  together ; 
aXbXc 


then 


P 


a'  XUXd 


Q.  E.  D. 


298 


GEOMETRY.  —  BOOK   VII. 


Proposition  X.     Theorem. 
538.  The  volume  of  a  rectangular  parallelopiped  is  equal 
to  the  product  of  its  three  dimensions,  the  unit  of  volume  being 
a  cube  whose  edge  is  the  linear  unit. 


/ 

'  /  /  /  / 

/  /  /  /  2 

/  / 

/ 
/ 

/ 

/ 

/ 

t  A 

y 

1 


Let  a,  b,  and  c  be  the  three  dimensions  of  the  rectan- 
gular parallelopiped  P,  and  let  the  cube  U  be  the 
unit  of  volume. 

We  are  to  prove       volume  of  P  =  aXbX  c. 
P       aXbXc 
U~ 


1X1X1 
p 

But  _  is  the  volume  of  P ; 

.'.  the  volume  of  P  =  a  X  b  X  c. 


§537 
§500 


Q.  E.  D. 


539.  Corollary  I.  Since  a  cube  is  a  rectangular  parallelo- 
piped having  its  three  dimensions  equal,  the  volume  of  a  cube  is 
equal  to  the  third  power  of  its  edge. 

540.  Cor.  II.  The  product  a  X  b  represents  the  base  when 
c  is  the  altitude  ;  hence  :  The  volume  of  a  rectangular  parallelo- 
piped is  equal  to  the  product  of  its  base  by  its  altitude. 

541.  Scholium.  When  the  three  dimensions  of  the  rec- 
tangular parallelopiped  are  each  exactly  divisible  by  the  linear 
unit,  this  proposition  is  rendered  evident  by  dividing  the  solid 
into  cubes,  each  equal  to  the  unit  of  volume.  Thus,  if  the  three 
edges  which  meet  at  a  common  vertex  contain  the  linear  unit 
3,  4  and  5  times  respectively,  planes  passed  through  the  several 
points  of  division  of  the  edges,  and  perpendicular  to  them,  will 
divide  the  solid  into  cubes,  each  equal  to  the  unit  of  volume ; 
and  there  will  evidently  be  3  X  4  X  5  of  these  cubes. 


PRISMS. 


299 


Proposition  XL     Theorem. 
542.    The  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  by  its  altitude. 


KB  J 

Let  A  BC  D-F  be  a  parallelopiped  having  all  its  faces 
oblique,  and  HE  its  altitude. 

We  are  to  prove  A  B  G  D-F  =ABGDX  HE. 

By  making  the  right  section  HUN  and  completing  the 

parallelopiped  HIJ  N-GLKM  we  have  a  right  parallelopiped 

equivalent  to,  A  B  G  D-F.  §  529 

(an  oblique  prism  is  equivalent  to  a  right  prism  whose  base  is  a  right  section 

of  the  oblique  prism  and  whose  altitude  is  equal  to  a  lateral  edge  of 

the  oblique  jJrism). 

Through  the  edge  I L  make  the  right  section  ILPO,  and 
complete  the  right  parallelopiped  ILPO-HGQE,  and  we  have 
a  rectangular  parallelopiped  equivalent  to  H I J  N-G  LKM>\  529 

and  hence  equivalent  to  A  B  G  D-F. 

Kow  O  ILGH^O  FFGH, 

O  0PQE  =  (OILGH)  =  OJKMN; 

and  O  ABGD  =  0  EFGH. 

.'.n  OPQE-O  ABCD. 

Moreover,  the  three  parallelopipeds  have  the  common  alti- 
tude HE. 

But        OPQE-ILGH=GPQEX  HE;  §540 

.\ABGD-F=ABGD  X  HE. 

Q.  E.  D. 


§322 
§530 
§530 


300 


GEOMETRY. BOOK  VII. 


Proposition  XII.     Theorem. 
543.    The  volume  of  any  prism  is  equal  to  the  product 

of  its  base  by  its  altitude. 
Ef 


Case  I.  —  When  the  base  is  a  triangle. 
Let    V  denote    the   volume,  B  the   base,   and  H  the 
altitude  of  the  triangular  prism  A  EC-E'. 
We  are  to  prove  V  =  B  X  H. 

Upon  the  edges  A  E,  EC,  E E',  construct  parallelopiped 
AEG  D-E'. 

Then  A  E  C-E>  =s=  \  A  E  G  D-E',  §  532 

(the  plane  passed  through  two  diagonally  opposite  edges  of  a  parallelopiped 
divides  it  into  two  equivalent  triangular  prisms), 

and  AEC=\AEGD.  §133 

But  A  E  CD-E'  =  2BX  H,  §  542 

(the  volume  of  any  parallelopiped  is  equal  to  the  product  of  its  base  by  its 

altitude). 

.'.  V=  J  (2  BX  H)  =  BX  H. 

Case  II.  — -  When  the  base  is  a  polygon  of  more  than  three  sides. 

Planes  passed  through  the  lateral  edge  A  A',  and  the  several 
diagonals  of  the  base  will  divide  the  given  prism  into  triangular 
prisms, 

which  have  for  a  common  altitude  the  altitude  of  the  prism. 

Hence,  the  volume  of  the  entire  prism  is  the  product  of  the 
sum  of  their  bases  by  the  common  altitude , 

that  is  the  entire  base  by  the  altitude  of  the  prism. 

Q.  E.  D. 

544.  Corollary.  Prisms  having  equivalent  bases  are  to 
each  other  as  their  altitudes ;  prisms  having  equal  altitudes  are 
to  each  other  as  their  bases ;  and  any  two  prisms  are  to  each 
other  as  the  product  of  their  bases  and  altitudes.  Any  two 
prisms  having  equivalent  bases  and  equal  altitudes  are  equivalent. 


PRISMS.  301 

Proposition  XIII.     Theorem. 
545.    The  four  diagonals  of  a  parallelopiped  bisect  each 

n 


Let  AG,  EC,  B H,  and  FD%  be  the  four  diagonals  of 
the  parallelopiped  A  G. 
We  are  to  prove  these  four  diagonals  bisect  each  other. 

Through  the  opposite  and  II  edges  A  E  and  C  G  pass  a  plane 
intersecting  the  II  bases  in  the  II  lines  A  C  and  E  G. 

The  section  A  C  G  E  is  a  O, 

(having  Us  op})ositc  sides  II ) ; 

.'.  its  diagonals  AG  and  EC  bisect  each  other  in  the 
point  0.  §  138 

In  like  manner  a  plane  passed  through  the  opposite  and  II 
edges  FG  and  A  D  will  form  a  O  AFGD, 

whose  diagonals  A  G  and  F  D  will  bisect  each  other  in  the 
point  0.  §  138 

Also,  a  plane  passed  through  the  opposite  and  II  edges  E  H 
and  B  C  will  form  a  O  E  B  G  H, 

whose  diagonals  E  C  and  B  H  will  bisect  each  other  in  the 
point  0. 

,\  the  four  diagonals  bisect  each  other  at  the  point  0. 

Q.  E.  D. 

546.  Corollary.  The  diagonals  of  a  rectangular  parallelo- 
piped are  equal. 

547.  Scholium.  The  point  0,  in  which  the  four  diagonals 
intersect,  is  called  the  centre  of  the  parallelopiped  ;  and  it  is  evi- 
dent that  any  straight  line  drawn  through  the  point  0  and 
terminated  by  two  opposite  faces  of  the  parallelopiped  is  bisected 
at  that  point.     Hence  0  is  the  centre  of  symmetry. 


802 


GEOMETRY. BOOK    VII. 


On  Pyramids. 

548.  Def.  A  Pyramid  is  a  polyhedron  one  of  whose  faces 
is  a  polygon,  and  whose  other  faces  are  triangles  having  a  com- 
mon vertex  and  the  sides  of  the  polygon  for  bases. 

549.  Def.  The  Base  of  a  pyramid  is  the  face  whose  sides 
are  the  bases  of  the  triangles  having  a  common  vertex. 

550.  Def.  The  Lateral  faces  of  a  pyramid  are  all  the  faces 
except  the  base. 

551.  Def.  The  Lateral  surface  of  a  pyramid  is  the  sum  of 
its  lateral  faces. 

552.  Def.  The  Lateral  edges  of  a  pyramid  are  the  intersec- 
tions of  its  lateral  faces. 

553.  Def.  The  Basal  edges  of  a  pyramid  are  the  intersec- 
tions of  its  base  with  its  lateral  faces. 

554.  Def.  The  Vertex  of  a  pyramid  is  the  common  vertex 
of  its  lateral  faces. 


555.  Def.  The  Altitude  of  a 
pyramid  is  the  perpendicular  distance 
from  its  vertex  to  the  plane  of  its 


Thus,  V-ABCDE is  a  pyramid ; 
ABCDEis  its  base;  AVB,BVC, 
etc.  are  its  lateral  faces,  and  their  sum    ^ 
is  its  lateral  surface;   V A,  V B,  etc. 
are  its  lateral  edges ;  A  B,  B  C,  etc.  @  B 

its  basal  edges ;   V  is  its  vertex  ;  V  0  is  its  altitude. 


PYRAMIDS.  303 

556.  Def.  A  Regular  pyramid  is  a  pyramid  whose  base  is 
a  regular  polygon,  and  whose  vertex  is  in  the  perpendicular  to 
the  base  at  its  centre. 

557.  Def.  The  Axis  of  a  regular  pyramid  is  the  straight 
line  joining  its  vertex  with  the  centre  of  the  base. 

558.  Def.  The  Slant  height  of  a  regular  pyramid  is  the 
altitude  of  any  lateral  face. 

559.  Def.  A  pyramid  is  triangular,  quadrangular,  pentag- 
onal,  etc.,  according  as  its  base  is  a  triangle,  quadrilateral, 
pentagon,  etc.  A  triangular  pyramid  formed  by  four  faces  (all 
of  which  are  triangles)  is  a  tetrahedron. 

560.  Def.  A  Truncated  pyramid  is 
the  portion  of  a  pyramid  included  be- 
tween its  base  and  a  section  cutting  all 
its  lateral  edges. 

561.  Def.  A  Frustum  of  a  pyramid 
is  a  truncated  pyramid  in  which  the  cut- 
ting section  is  parallel  to  the  base. 

562.  Def.   The  base  of  the  pyramid 

is  called  the  Lower  base  of  the  frustum,  and  the  parallel  sec- 
tion, its  Upper  base. 

563.  Def.  The  Altitude  of  a  frustum  is  the  perpendicular 
distance  between  the  planes  of  its  bases. 

564.  Def.  The  lateral  faces  of  a  frustum  of  a  regular  pyra- 
mid are  the  trapezoids  included  between  its  bases ;  the  lateral 
surface  is  the  sum  of  the  lateral  faces;  the  Slant  height  of  a 
frustum  of  a  regular  pyramid  is  the  altitude  of  any  lateral  face. 


304 


GEOMETRY.  —  BOOK    VII. 


Proposition  XIV.     Theorem. 

565.   If  a  pyramid  be  cnt  by  a  plane  parallel  to  its  base, 
I.   The  edges  and  altitude  are  divided  proportionally  j 
II.  The  section  is  a  polygon  similar  to  the  base. 

vi  y 


Let  the  pyramid  V-A  B  0  D  E,  whose  altitude  is  V  0, 
be  cut  by  a  plane  abode  parallel  to  its  base,  in- 
tersecting the  lateral  edges  in  the  points  a,  b,  c,  d,  e, 
and  the  altitude  in  o. 

We  are  to  prove 
I  Va__     Vb  Vo  . 

VA~  VB  VO* 

II.    The  section  abcde  similar  to  the  base  ABODE. 

I.  Suppose  a  plane  passed  through  the  vertex  V  II  to  the  base. 

Then  the  edges  and  the  altitude  will  be  intersected  by  three 
II  planes. 

.    Va_       Vb_  Vo_ 

"  VA~  VB  VO' 

{if  straight  lines  be  intersected  by  three  II  planes,  their  corresponding  segment* 
are  -proportional). 

II.  The  sides  ab,  be  etc.  are  parallel  respectively  to  A  B,  B  C, 

etc.,  §  465 

(the  intersections  of\\  planes  by  a  third  plane  are  II  lines)  ; 
.'.  A  a  be,  bed  etc.  are  equal  respectively   to  A  ABC, 

BOD  etc.,  §462 

(if  two  A  not  in  the  same  plane  have  their  sides  respectively  II  and  lying  in 
the  same  direction,  they  are  equal). 

.*.  the  two  polygons  are  mutually  equiangular. 


§469 


PYRAMIDS.  305 

Also,  since  the  sides  of  the  section  are  II  to  the  correspond- 
ing sides  of  the  base, 

A  Vab,  Vbc  etc.  are  similar  respectively  to  A  VA  B, 
VB  C  etc. 

... Jtl  -(™Xi±L=(I±)  =  l±  etc. 

AB       \VB/~  BC~  \VCJ~  CD 

.'.the  polygons  have  their  homologous  sides  proportional; 
.*.  section  a b c d e  is  similar  to  the  base  ABC D E.      §  278 

Q.  E.  D. 

566.  Corollary  1.  Any  section  of  a  pyramid,  parallel  to 
its  base  is  to  the  base  as  the  square  of  its  distance  from  the  ver- 
tex is  to  the  square  of  the  altitude  of  the  pyramid. 

Since  £»       (™)_£i. 

VO       \VB/       AB 

Squaring  TW=TW- 

a  b  c  d  e  a  b 

Eut  ABCDE=  TJ?'  §344 

(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 

a  b  c  d  e         V  o 


"ABCDE      Yd2' 

567.  Cor.  2.  If  two  pyramids  having  equal  altitudes  be  cut 
by  planes  parallel  to  their  bases,  and  at  equal  distances  from 
their  vertices,  the  sections  will  have  the  same  ratio  as  their  bases. 

For 


and 


Now,  since  Vo  =  V o1 ,  and  VO  =  V  0', 

abode  :  ABCDE  :  -.a'b'd  :  A1  B'  C. 
Whenceabcde  \a'b'd  :  :  AB  C D  E  :  A' B' C.     §  262 
568.  Cor.  3.    If  two  pyramids  have  equal  altitudes   and 
equivalent  bases,  sections  made  by  planes  parallel  to  their  bases 
and  at  equal  distances  from  their  vertices  are  equivalent. 


abode 

To2 

ABCDE 

VO2' 

a'V  d 

Vo* 

A'B'C1 

VO*' 

306 


GEOMETRY.  —  BOOK    VII. 


Proposition  XV.     Theorem. 
569.    The  lateral  area  of  a  regular  ^ 
one-half  the  product  of  the  perimeter  of  its 
height.  v 


is  equal  to 
by  its  slant 


Let   V-ABCDE  be  a  regular  pyramid,  and   VH  its 
slant  height. 

We  are  to  prove  the  sum  of  the  faces  V  AB,  V  BG,  etc.  =  J 
(AB  +  BO,  etc.)  X  VH. 

Now  AB=BC=  CD,  etc.,  §363 

(being  sides  of  a  regular  polygon). 

VA=  VB  =  VC,  etc.,  §  450 

(oblique  lines  dravm  from  any  point  in  a  ±  to  a  plane  at  equal  distances 
from  the  foot  of  the  A.  are  equal). 

.'.  A  VAB,  VBG,  etc.  are  equal  isosceles  A,      §  108 

whose  bases  are  the  sides  of  the  regular  polygon  and  whose 
common  altitude  is  the  slant  height  VH. 

Now  the  area  of  one  of  these  A,  as  VAB,=  ^  base  AB  X 
altitude  VH,  §  324 

.*.  the  sum  of  the  areas  of  these  A,  that  is,  the  lateral  area 
of  the  pyramid,  is  equal  to  J  the  sum  of  their  bases 
(AB  +  BC  +  CD,  etc.)  X  V H. 

Q.  E.  D. 

570.  Corollary  1.  The  lateral  area  of  the  frustum  of  a 
regular  pyramid,  being  composed  of  trapezoids  which  have  for 
their  common  altitude  the  slant  height  of  the  frustum,  is  equal  to 
one-half  the  sum  of  the  perimeters  of  the  bases  multiplied  by  the 
slant  height  of  the  frustum. 

571.  Cor.  2.  The  dihedral  angles  formed  by  the  intersec- 
tions of  the  lateral  faces  of  a  regular  pyramid  are  all  equal.  §  492 


PYRAMIDS. 


307 


Proposition  XYI.     Theorem. 
572.    Two  triangular  pyramids  having  equivalent  bases 
and  equal  altitudes  are  equivalent. 
X 


Let  S-ABC  and  S'-A'  B'  C  be  two  triangular  pyramids 
having  equivalent  bases  ABC  and  A'  B'C  situated 
in  the  same  plane,  and  a  common  altitude  A  X. 
We  are  to  prove       S-ABC  ^  S'-A'  B'  C. 
Divide  the  altitude  A  X  into  a  number  of  equal  parts, 
and  through  the  points  of  division  pass  planes  II  to  the 

planes  of  their  bases,  intersecting  the  two  pyramids. 

In  the  pyramids  S-ABC  and  S'-A' B' C  inscribe  prisms 

whose  upper  bases  are  the  sections  D  E  F,  G  U I,  etc.,  D'  E'  F', 

G'HT,  etc. 

The  corresponding  sections  are  equivalent,  §  568 

(if  two  pyramids  have  equal  altitudes  and  equivalent  bases,  sections  made  by 
planes  II  to  their  bases  and  at  equal  distances  from  their  vertices  are 
equivalent). 

.'.  the  corresponding  prisms  are  equivalent,  §  544 

(prisms  having  equivalent  bases  and  equal  altitudes  are  equivalent). 

Denote  the  sum  of  the  prisms  inscribed  in  the  pyramid 
S-A  B  C,  and  the  sum  of  the  corresponding  prisms  inscribed  in 
the  pyramid  S'-A'  B'  C  by  V  and  V  respectively. 

Then  F=  V. 

Now  let  the  number  of  equal  parts  into  which  the  altitude 
A  X  is  divided  be  indefinitely  increased  ; 

The  volumes  V  and   V  are  always  equal,  and  approach  to 

the  pyramids  S-A  B  C  and  S'-A'  B'  C  respectively  as  their  limits. 

Hence  S-A  B  C  o  S'-A'  B'  C.  §  1 99 

Q.  E.  D. 


308  GEOMETEY. BOOK   VII. 

Proposition  XVII.     Theorem. 
573.    The  volume  of  a  triangular  pyramid  is  equal  to  one- 
third  of  the  product  of  its  base  and  altitude. 


Let  S-ABC be  a  triangular  pyramid,  and  H its  altitude. 

We  are  to  prove  S-A  B  G  «=  J  A  B  G  X  H. 

On  the  base  ABC  construct  a  prism  ABG-SED,  having  its 
lateral  edges  II  to  SB  and  its  altitude  equal  to  that  of  the  pyramid. 

The  prism  will  be  composed  of  the  triangular  pyramid 
S-A  B  G  and  the  quadrangular  pyramid  S-A  G  D  E. 

Through  S  A  and  S  D  pass  a  plane  SAD. 

This  plane  divides  the  quadrangular  pyramid  into  the  two 
triangular  pyramids,  S-A  G  D  and  S-A  E  D ,  which  have  the  same 
altitude  and  equal  bases.  §  133 

r.S-AG D=o=  S-A  ED,  §572 

(two  triangular  pyramids  having  equivalent  bases  and  equal  altitudes  are 
equivalent). 

Now  the  pyramid  S-A  E  D  may  be  regarded  as  having 
ESD  for  its  base  and  A  for  its  vertex. 

.'.  pyramid  S-A  E D  =©=  pyramid  S-A  BG,  §  572 

(having  equal  bases  SED  and  ABC  and  the  same  altitude). 

.'.  the  three  pyramids  into  which  the  prism  A  B  G-SE D  is 
divided  are  equivalent. 

.*.  pyramid  S-A  B  G  is  equivalent  to  J  of  the  prism. 

But  the  volume  of  the  prism  is  equal  to  the  product  of  its 
base  and  altitude  ;  §  543 

.*.  S-ABG  =  iABCX  H. 

Q.  E.  D. 


PYRAMIDS.  309 


Proposition  XVIII.     Theorem. 
574.    The  volume  of  any  pyramid  is  equal  to  one-third 
the  product  of  its  base  and  altitude. 


Let  S-A  BC  D  E  be  any  pyramid. 

We  are  to  prove  S-A  B  C  D  E  =  \ABCDEXSO. 

Through  the  edge  SB,  and  the  diagonals  of  the  base  DA, 
D  B,  pass  planes. 

These  divide  the  pyramid  into  triangular  pyramids,  whose 
bases  are  the  triangles  which  compose  the  base  of  the  pyramid, 

and  whose  common  altitude  is  the  altitude  SO  of  the 
pyramid. 

The  volume  of  the  given  pyramid  is  equal  to  the  sum  of  the 
volumes  of  the  triangular  pyramids. 

But  the  sum  of  the  volumes  of  the  triangular  pyramids  is 

equal  to  \  the  sum  of  their  bases  multiplied  by  their  common 

altitude,  §  573 

{the  volume  of  a  triangular  pyramid  is  equal  to  one-third  the  product  of  its 

base  and  altitude), 

that  is,  the  volume  of  the  pyramid  S-A  BC D E  =  J 
ABCDE  X  SO. 

Q.  E.  D. 

575.  Corollary.  Pyramids  having  equivalent  bases  are  to 
each  other  as  their  altitudes  ;  pyramids  having  equal  altitudes 
are  to  each  other  as  their  bases.  Any  two  pyramids  are  to  each 
other  as  the  products  of  their  bases  and  altitudes. 

576.  Scholium.  The  volume  of  any  polyhedron  may  be 
found  by  dividing  it  into  pyramids,  and  computing  the  volumes 
of  these  pyramids  separately. 


310 


GEOMETRY.  —  BOOK    VII. 


Proposition  XIX.     Theorem. 

577.    Two  tetrahedrons  having  a  trihedral  angle  of  the 
one  equal  to  a  trihedral  angle  of  the  other  are  to  each  other  as 
the  products  of  the  three  edges  of  these  trihedral  angles. 
V. 


Let  V  and  V  denote  the  volumes  of  the  two  tetra- 
hedrons D-ABC,  D'-AB'C,  having  the  trihedral  A 
of  the  one  equal  to  the  trihedral  A  of  the  other. 

^         t  V       AB X  ACX AD 

We  are  to  prove    —  = 

r  V       AB'XAC'X  AD' 

Place  the  tetrahedrons  so  that  their  equal  trihedral  A  shall 
be  in  coincidence. 

Consider  ABC  and  A  B'  C  the  bases  of  the  two  tetrahe- 
drons, 

and  from  D  and  D1  draw  D  0  and  D'  0'  J_  to  the  base  ABO. 


Now 


ABC  X  DO         ABO        DO 
AB'C'X  D'  0'  ~  AB'C'X  D<  O1 


§  575 


{any  two  pyramids  are  to  each  other  as  the  products  of  their  bases  and 

altitudes). 


But 


and 


ABC        ABX AG 


AB'  C       AB'XAC 
DO       AD 


D'  0'      AD' 

{being  homologous  sides  of  the  similar  &.ADO  and  A  D1 0'). 


§341 
§  278 


V 


ABX  ACX  AD 
A  B'  X  ACX  AD' 


Q.  E.  D. 


exercises.  811 

Exercises. 

1.  Given  a  cubical  tank  holding  one  ton  of  water ;  find  its 
length  in  feet,  if  a  cubic  foot  of  water  weigh  1000  ounces. 

2.  At  17  cents  a  square  foot,  what  is  the  cost  of  lining  with 
zinc  a  rectangular  cistern  5  ft.  7  in.  long,  3  ft.  11  in.  broad,  2  ft. 

8  J  in.  deep  1 

3.  Find  the  side  of  a  cubical  block  of  cast  iron  weighing  a 
ton,  if  iron  weigh  7.2  as  much  as  water,  and  a  cubic  foot  of 
water  weigh  1000  ounces. 

4.  How  many  cubic  yards  of  gravel  will  be  required  for  a 
walk  surrounding  a  rectangular  lawn  200  yards  long,  and  100 
yards  wide ;  the  walk  to  be  3  feet  wide  and  the  gravel  3  inches 
deep] 

5.  The  volume  of  a  rectangular  solid  is  the  sum  of  two  cubes 
whose  edges  are  10  inches  and  2  inches  respectively,  and  the 
area  of  its  base  is  the  difference  between  2  squares  whose  sides 
are  1£  feet  and  1£  feet  respectively ;  find  its  altitude  in  feet. 

6.  A  rectangular  cistern  whose  length  is  equal  to  its  breadth  is 
22  decimetres  deep,  and  contains  10  tonneaux  of  water;  find  its 
length. 

7.  Given  a  regular  prism  whose  base  is  a  regular  hexagon  in- 
scribed in  a  circle  6  metres  in  diameter,  and  whose  altitude  is 
8.7  metres ;  find  the  number  of  kilolitres  it  will  contain,  if  the 
thickness  of  the  walls  be  1  decimetre. 

8.  A  pond  whose  area  is  11  hectares,  21  ares,  22.2  centares, 
is  covered  with  ice  21  centimetres  thick.  What  is  the  weight  of 
this  body  of  ice  in  kilogrammes,  the  weight  of  ice  being  92  % 
that  of  water. 

9.  Given  two  hollow  oblique  prisms,  whose  interior  dimen- 
sions are  as  follows  :  the  area  of  a  right  section  of  the  first  is  18 
sq.  ft.,  of  the  second  2.1  sq.  metres ;  a  lateral  edge  of  the  first  is 

9  ft.,  of  the  second  2.1  metres ;  find  the  volume  of  each  in  cubic 
yards,  cubic  metres,  cubic  decimetres,  and  cubic  centimetres; 
find  the  capacity  of  each  in  gallons  and  litres,  in  bushels  and 
hectolitres  ;  and  find  the  weight  of  water  in  pounds  and  in  kilo- 
grammes, required  to  fill  each  prism. 


312 


GEOMETRY. BOOK   VII. 


Proposition  XX.     Theorem. 


578.  The  frustum  of  a  triangular  pyramid  is  equivalent 
to  the  sum  of  three  pyramids  whose  common  altitude  is  the 
altitude  of  the  frustum  and,  whose  bases  are  the  lower  base, 
the  tipper  base,  and  a  mean  proportional  between  the  two  bases 
of  the  frustum. 

J 


Let  B  and  b  denote  the  lower  and  upper  bases  of  the 
frustum  ABC-DEF,  and  H  its  altitude. 

Through  the  vertices  A,  E,  G  and  E,  D,  G  pass  planes 
dividing  the  frustum  into  three  pyramids. 

Now  the  pyramid  E-A  B  G  has  for  its  altitude  H,  the  alti- 
tude of  the  frustum,  and  for  its  base  B,  the  lower  base  of  the 
frustum. 

And  the  pyramid  C-E  D  F  has  for  its  altitude  H,  the  alti- 
tude of  the  frustum,  and  for  its  base  b,  the  upper  base  of  the 
frustum.     Hence,  it  only  remains 

To  prove  E-A  D  G  equivalent  to  a  pyramid,  having  for  its 
altitude  H,  and  for  its  base  \B  X  b. 

E-A  B  C  and  E-A  D  G,  regarded  as  having  the  common  ver- 
tex G,  and  their  bases  in  the  same  plane  B  D,  have  a  common 
altitude. 

.'.  E-A  B  G  :  E-A  D  G  :  :  A  A  E B  :  A  A  E D.      §  575 

(pyramids  having  equal  altitudes  are  to  each  other  as  their  bases). 

Now  since  the  AAEB  and  A  E D  have  a  common  altitude, 
(that  is,  the  altitude  of  the  trapezoid  A  BED), 


we  have     AAEB  :  A  AED  :  :AB  :  D  E, 


326 


PYRAMIDS.  313 

.-.E-ABC  :  E-A  DC  :  :  A  B  :  D  E. 

In  like  manner  E-A  D  C  and  E-D  F  C,  regarded  as  having 
the  common  vertex  E  and  their  bases  in  the  same  plane  D  C, 
have  a  common  altitude. 

.'.E-A  DC  :  E-DFC  ::AADC  :A  DEC.      §  575 

But  since  the  A  A  D  C  and  DEC  have  a  common  altitude, 
(the  altitude  of  the  trapezoid  A  CF  D), 

we  have      A  A  D  C  :  A  D  EC  :  :  A  C  :  D F.  §326 

Now  A  D  E  F  is  similar  to  A  A  B  C,  §  565 

(the  section  of  a  pyramid  made  by  a  plane  II  to  the  base  is  a  polygon  similar 
to  the  base)  ; 

.'.AB  :DE  :  :  A  C  :  D F.  §  278 

.'.E-ABC  :  E-A  DC  :  :  E-A  DC  :  E-DFC. 
Now  E-ABC  =  i  HX  B,  §  573 

and  E-DFC =  C-E D F  =  \  H  X  b.  §573 

.'.  E-A  DC  =  \J%HXBX%HXb  =  J  #  ^  X  6. 

Q.  E.  D. 

579.  Corollary  1.  Since  the  volume  of  the  frustum  is  de- 
noted by  V,  the  lower  base  by  B,  the  upper  base  by  b,  and  the 
altitude  by  Hy 

we  have  V=\HXB+\HXb  +  \HX  ^Wx~b 
=  $HX(B+b+  \jTx~b). 

580.  Cor.  2.  The  frustum  of  any  pyramid  is  equivalent  to 
the  sum  of  three  pyramids  whose  common  altitude  is  the  altitude 
of  the  frustum,  and  whose  bases  are  the  loiver  base,  the  upper  base, 
and  a  mean  proportional  between  the  bases  of  the  frustum. 

For  the  frustum  of  any  pyramid  is  equivalent  to  the  corre- 
sponding frustum  of  a  triangular  pyramid  having  the  same  alti- 
tude and  an  equivalent  base  (§  578) ;  and  the  bases  of  the  frustum 
of  a  triangular  pyramid  being  both  equivalent  to  the  correspond- 
ing bases  of  the  given  frustum,  a  mean  proportional  between  the 
triangular  bases  is  equivalent  to  a  mean  proportional  between 
their  equivalents. 


314 


GEOMETRY. BOOK    VII. 


Proposition  XXL     Theorem. 

581.  A  truncated  triangular  prism  is  equivalent  to  the 
sum  of  three  pyramids  whose  common  base  is  the  base  of  the 
prism,  and  whose  vertices  are  the  three  vertices  of  the  inclined 
section. 


Let  AB  C-D  E F  be  a  truncated  triangular  prism  whose 
base  is  ABC,  and  inclined  section  D  E  F. 

We  are  to  prove  A  B  C-D  E  F  =0=  three  pyramids,  E-A  B  G, 
D-A  B  C  and  F-A  B  G. 

Pass  the  planes  AEG  and  DEC,  dividing  the  truncated 
prism  into  the  three  pyramids  E-A  B G,  E-A  G D,  and  EG D F. 

Now  the  pyramid  E-A  B  G  has  the  base  ABC  and  the 
vertex  E. 

E-AGDoB-ACD,  §574 

(for  they  have  the  same  base  AC  D  and  the  same  altitude,  since  their  vertices 
E  and  B  are  in  the  line  EB  II  to  the  base  A  CD), 

But  pyramid  B-A  CD,  which  is  equivalent  to  pyramid 
E-A  C I),  may  be  regarded  as  having  the  base  ABC  and  the 
vertex  D. 

Again,  E-CDF*>  B-A  C  F, 

for  their  bases  CDF  and  AG F,  in  the  same  plane,  are 

equivalent,  §  325 

{for  the  A  CDF  and  A  CFhave  the  common  base  C F  and  equal  altitudes, 
their  vertices  lying  in  the  line  A  D\\  to  C F). 


PYRAMIDS. 


315 


Moreover,  E-C D  F  and  B-A  C  F  have  the  same  altitude, 

(since  their  vertices  E  and  B  are  in  the  line  E  B  II  to  the  plane  of  their 
bases  A  CDF). 

But  the  pyramid  B-A  C  F  may  be  regarded  as  having  the 
base  ABC  and  the  vertex  F. 

.'.the  truncated  triangular  prism  A  B  C-DEFis  equivalent 
to  the  three  pyramids  E-A  B  C,  DA  B  C,  and  F-A  B  C. 

Q.  E.  D 
F 


582.  Corollary  1.  The  volume  of  a  truncated  right  tri 
angular  prism  is  equal  to  the  product  of  its  base  by  one-third 
the  sum  of  its  lateral  edges.  For  the  lateral  edges  D  A>  EB, 
FC,  being  perpendicular  to  the  base,  are  the  altitudes  of  the 
three  pyramids  whose  sum  is  equivalent  to  the  truncated  prism. 
And,  since  the  volume  of  a  pyramid  is  one-third  the  product  of 
its  base  by  its  altitude,  the  sum  of  the  volumes  of  these  pyramids 
=  ABCXi(DA  +  EB  +  FC). 

583.  Cor.  2.  The  volume  of  any  truncated  triangular  prism 
is  equal  to  the  product  of  its  right  section  by  one-third  the  sum 
of  its  lateral  edges. 

For  let  A  B  C-Al  B'  C  be  any  truncated  triangular  prism. 
Then  the  right  section  D  E  F  divides  it  into  two  truncated  right 
prisms  whose  volumes  are  D  E  F  X  J  (A  D  +  B  E  +  C  F)  and 
DEFX  £  (A'D  +  B'E-b  C F). 

Whence  their  sum  is  D  EF  X  J  (A  A'  +  B  B'  +  C  C). 


316  GEOMETRY.  —  BOOK   VIE. 


Exercises. 

1.  Given  a  pyramid  whose  base  is  a  rectangle  80  feet  by  60 
feet,  and  whose  lateral  edges  are  each  1 30  feet ;  find  its  volume, 
and  its  entire  surface. 

2.  Given  the  frustum  of  a  pyramid  whose  bases  are  hepta- 
gons ;  each  side  of  the  lower  base  being  10  feet,  and  of  the  upper 
base  6  feet,  and  the  slant  height  42  feet ;  find  the  convex  surface 
in  square  yards. 

3.  Given  a  stick  of  timber  30  feet  long,  the  greater  end  being 
18  inches  square,  and  the  smaller  end  15  inches  square;  find  its 
volume  in  cubic  feet. 

4.  Given  a  stone  obelisk  in  the  form  of  a  regular  quadrangular 
pyramid,  having  a  side  of  its  base  equal  to  25  decimetres,  and  its 
slant  height  12  metres.  The  stone  weighs  2.5  as  much  as  water. 
What  is  its  weight  in  kilogrammes  1 

5.  Given  the  frustum  of  a  pyramid  whose  bases  are  squares  j 
each  side  of  the  lower  base  being  35  decimetres,  each  side  of  the 
upper  base  25  decimetres,  and  the  altitude  15  metres ;  find  its 
volume  in  steres. 

6.  Given  a  right  hexagonal  pyramid  whose  base  is  inscribed 
in  a  circle  30  feet  in  diameter,  and  whose  altitude  is  20  feet ; 
find  its  convex  surface,  and  its  volume. 

7.  Given  a  right  pentagonal  pyramid  whose  base  is  inscribed 
in  a  circle  20  feet  in  diameter,  and  whose  slant  height  is  30  feet ; 
find  its  convex  surface,  and  its  volume. 

8.  Find  the  difference  between  the  volume  of  -the  frustum  of 
a  pyramid,  and  the  volume  of  a  prism  of  the  same  altitude  whose 
base  is  a  section  of  the  frustum  parallel  to  its  bases  and  equidis- 
tant from  them. 

9.  Given  a  stick  of  timber  32  feet  long,  18  inches  wide,  15 
inches  thick  at  one  end,  and  12  inches  at  the  other;  find  the 
number  of  cubic  feet,  and  the  number  of  feet  board  measure  it 
contains.     Find  equivalents  for  the  results  in  the  metric  system. 


SIMILAR   POLYHEDRONS. 


317 


On  Similar  Polyhedrons. 

584.  Def.  Similar  polyhedrons  are  polyhedrons  which 
have  the  same  form.  They  have,  therefore,  the  same  number  of 
faces,  respectively  similar  and  similarly  placed,  and  their  corre- 
sponding polyhedral  angles  equal. 

585.  Def.    Homologous  faces,  lines,  and  angles  of  similar 

polyhedrons  are  faces,  lines,  and  angles  similarly  placed. 

8 


I.    The  homologous  edges  of  similar  polyhedrons  are  pro- 
portional. 

Since  the  faces  SAB,  SA  C,  SB  C  and  A  B  C  are  similar 
respectively  to  S'  A'  B',  S1  A'  C,  S'  B1  C  and  A'  B'  C,  we  have 


SA    _SB^_^    etc 
~SrA'~SrB'~ArB'1  6C* 


§278 


II.    Any  two  homologous  faces  of   similar  polyhedrons  are 
proportional  to  the  squares  of  any  two  homologous  edges. 


SAB 

Thus'  S'A'B' 


S  A2 


S'  A'2 


SAC        SC2       SBC 

S'A/C'~SrC<2~S'B'C''  §342 


III.    The  entire  surfaces  of  two  similar  polyhedrons  are  pro- 
portional to  the  squares  of  any  two  homologous  edges. 


™         .  SAB        SAC       . 

Inus,  since       . ,  = ,  etc., 

S'A'B'      S'A'C 


SAB  +  SAC,  etc.        SAB       -$-# 


S'  A'  B'  +  S'  A'  C,  etc.      S'  A'  B'      ^p  ' 


266 


318 


GEOMETRY. 


BOOK    VII. 


Proposition  XXII.     Theorem. 

586.  Two  similar  polyhedrons  may  be  decomposed  into 
the  same  number  of  tetrahedrons  similar,  each  to  each,  and 
similarly  placed. 


Let  ABCDEOPQRS  and  A1  B' C D' E'-O' F  Q'  R' &  be 
two  similar  polyhedrons  of  which  P  and  P'  are 
homologous  vertices. 

We  are  to  prove  that  A  BCDE-OPQRS  and  A'B'G'D'E'- 
0'  P'  Q1  R'  S'  can  be  decomposed  into  the  same  number  of  tetrahe- 
drons similar  and  similarly  placed. 

Place  two  homologous  faces  A  BCD  and  A'B'O'D'  in 
the  same  plane,  having  two  homologous  edges  AB  and  A'  B'  II 
and  lying  in  the  same  direction. 

On  any  two  corresponding  faces  not  adjacent  to  P  and  P'> 
as  ABODE  and  A' B'  C I> E',  from  two  homologous  vertices, 
as  E  and  E',  draw  diagonals  dividing  these  faces  into  A,  similar 
and  similarly  placed. 

From  the  homologous  vertices  P,  P'  of  the  polyhedrons 
draw  straight  lines  to  the  vertices  of  these  A. 

Repeat  this  construction  for  each  of  the  faces  not  adjacent 
to  P,  P>. 

Then  the  polyhedrons  will  be  divided  into  the  same  number 
of  tetrahedrons ; 

that  is,  into  as  many  tetrahedrons  as  there  are  A  in  these 
faces. 


SIMILAR    POLYHEDRONS.  319 

Now,  any  two  corresponding  tetrahedrons,  as  P-A  B  E  and 
P'-A1  B'  E',  are  similar  ; 

for  the  faces  E  A  B  and  P  A  B  are  similar  respectively  to 
the  faces  E'  A'  B1  and  P'  A1  B',  §  294 

(being  similarly  situated  &  of  similar  polygons). 

In  the  A  PBE  and  P' B' E' 

PB\s\\  to  P'  B',  and  B  E  to  B' E't 

(since  they  make  equal  A  respectively  with  the  II  lines  A  B  and  A'  B') ; 

.\Z  PBE  =  Z  P'B'E't  §  462 

(two  A  not  in  the  same  plane  having  their  sides  II  and  lying  in  the  same 
direction  are  equal)  ; 

and  **=(A*\-M..  §278 

P'B>      \A'B'1      B'E'  * 

.*.  face  PBE  is  similar  to  face  P' B> E'.  §  284 

Also,  in  the  A  P  A  E  and  P'  A1  E' 

PE       (PB\       PA       (AB\       AE  .  27g 

P'E'~\P'B')      P'  A'~\A'  B')~  A'  E'*        S 
(being  homologous  sides  of  similar  A  ). 

.'.  face  P  A  E  is  similar  to  face  P'  A'  E'.  §  282 

Moreover,  since  any  two  corresponding  trihedral  A  of  these 

tetrahedrons  are  formed  by  three  plane  A  which  are  equal,  each 

to  each,  and  similarly  situated,  they  are  equal.  §  492 

.'.  P-A  BE  and  P'-A' B' E'  are  similar.  §  584 

In  like  manner  we  may  show  that  any  other  two  tetrahe- 
drons similarly  situated  are  similar. 

That  is,  the  two  similar  polyhedrons  have  the  same  number 
of  tetrahedrons  similar  each  to  each,  and  similarly  situated. 

Q.  E.  D. 

587.  Corollary.  Any  two  homologous  lines  in  two  similar 
polyhedrons  have  the  same  ratio  as  any  two  homologous  edges. 


320 


GEOMETRY. BOOK  VII. 


Proposition  XXIII.     Theorem. 
588.   Similar  tetrahedrons  are  to  each  other  as  the  cubes 
of  their  homologous  edges. 


Let  S-B  CD  and  S'-B'  CD'  be  two  similar  tetrahedrons 
having  for  bases  the  similar  faces  BCD  and  B1  CD', 
and  for  altitudes  S 0  and  S'  0'. 


We  are  to  prove 


SBC  D 


BCZ 
S'-B'C'D'~  W(j»' 


Apply  the  tetrahedron  S'-B'  C  D'  to  the  tetrahedron  S-B  C  D, 
so  that  the  polyhedral  S'  shall  coincide  with  S. 

Then  the  base  B' C D'  will  be  II  to  the  face  BCD, 
{since  their  planes  make  equal  A  with  the  face  SB  0), 

and  the  J_  S  0,  _L  to  B  C  D,  will  also  be  _L  to  B'  C  D'. 

SO'  will  be  the  altitude  of  the  tetrahedron  S-B' C D'. 


Now 


S-BCD        BCDXSO 


^x^,§575 
*  SO'    * 


S-B' CD'     B'C'D'XSO'     B' C  D' 

(any  two  tetrahedrons  are  to  each  other  as  the  products  of  their  bases  and 

altitudes). 


Since  the  bases  are  similar, 

BCD        B~& 


B'CD'      ^C1 


§343 


SIMILAR   POLYHEDRONS.  321 

Also,  = ,  §  587 

SO'      B'G'  * 

(in  two  similar  polyhedrons  any  two  homologous  lines  are  in  the  same  ratio 
as  any  two  homologous  edges). 

.    S-B  C  D       BC2       BC  _  BCZ 

s-B'c  iy~wc^x  B'tf^WW*' 

Q.  E.  D. 


589.  Corollary  1.  Two  similar  polyhedrons  are  to  each 
other  as  the  cubes  of  any  two  homologous  edges. 

For,  two  similar  polyhedrons  may  be  decomposed  into  tetra- 
hedrons similar,  eacli  to  each,  and  similarly  placed,  of  which  any 
two  homologous  edges  have  the  same  ratio  as  any  two  homolo- 
gous edges  of  the  polyhedrons.  And,  since  any  pair  of  the  simi- 
lar tetrahedrons  are  to  each  other  as  the  cubes  of  any  two 
homologous  edges,  the  entire  polyhedrons  are  to  each  other  as 
the  cubes  of  any  two  homologous  edges.  §  266 

590.  Cor.  2.  Similar  prisms  or  pyramids  are  to  each  other 
as  the  cubes  of  their  altitudes  ;  and  similar  polyhedrons  are  to  each 
other  as  the  cubes  of  any  two  homologous  lines. 


Ex.  1.  The  portion  of  a  tetrahedron  cut  off  by  a  plane  parallel 
to  any  face  is  a  tetrahedron  similar  to  the  given  tetrahedron. 

Ex.  2.  Two  tetrahedrons,  having  a  dihedral  angle  of  one  equal 
to  a  dihedral  angle  of  the  other,  and  the  faces  including  these 
angles  respectively  similar,  and  similarly  placed,  are  similar. 

Ex.  3.  Given  two  similar  polyhedrons,  whose  volumes  are  125 
feet  and  12.5  feet  respectively  j  find  the  ratio  of  two  homologous 
edges. 


322  GEOMETRY. BOOK   VII. 


On  Eegular  Polyhedrons. 

591.  Def.  A  Regular  polyhedron  is  a  polyhedron  all  of 
whose  faces  are  equal  regular  polygons,  and  all  of  whose  polyhe- 
dral angles  are  equal. 

The  regular  polyhedrons  are  the  tetrahedron,  octahedron  and 
icosahedron,  all  of  whose  faces  are  equal  equilateral  triangles; 
the  hexahedron,  or  cube,  whose  faces  are  squares ;  the  dodecahe- 
dron, whose  faces  are  regular  pentagons. 

Only  these  five  regular  polyhedrons  are  possible,  for  a  poly- 
hedral angle  must  have  at  least  three  face  angles,  and  must  have 
the  sum  of  its  face  angles  less  than  four  right  angles,  (§  488). 
Hence : 

I.  If  the  faces  be  equilateral  triangles,  polyhedral  angles 
may  be  formed  of  them  in  groups  of  3,  4,  or  5  only,  as  in  the 
tetrahedron,  octahedron  and  icosahedron.  Since  each  angle  of  an 
equilateral  triangle  is  two-thirds  of  a  right  angle,  the  sum  of  six 
such  angles  is  four  right  angles,  and  therefore  greater  than  a 
convex  polyhedral  angle. 

II.  If  the  faces  be  squares,  polyhedral  angles  may  be  formed 
of  them  in  groups  of  three  only,  as  in  the  regular  hexahedron,  or 
cube  ;  since  four  such  angles  would  be  four  right  angles. 

III.  If  the  faces  be  regular  pentagons,  polyhedral  angles 
may  be  formed  of  them  in  groups  of  three  only,  as  in  the  regular 
dodecahedron  ;  since  four  such  angles  would  be  greater  than  four 
right  angles. 

IV.  "We  can  proceed  no  farther ;  for  a  group  of  three  angles 
of  regular  hexagons  would  equal  four  right  angles,  and  of  regular 
heptagons,  etc.,  would  be  greater  than  four  right  angles. 


REGULAR   POLYHEDRONS. 


323 


450 


Proposition  XXIY.     Problem. 

592.    Given  an  edge,  to  construct  the  five  regular  poly- 
hedrons. 

Let  A  B  be  the  given  edge. 

I.    Upon  AB  to  construct  a  regular  tetrahedron. 

D 

Upon  A  B  construct  the  equilateral  A 
ABC.  §  232 

Find  the  centre  0  of  this  A,         §  238 
and  erect  0 D  _L  to  the  plane  ABC. 
Take  the  point  D  so  that  A  D  =  AB. 
Draw        DA,DB,DC. 
ABC D  is  the  regular  tetrahedron  required. 
For,  the  edges  are  all  equal, 
and  hence  the  faces  are  equal  equilateral  A. 
and  its  polyhedral  A  are  all  equal.  §  492 

q  II.  To  construct  a  regular  hexahedron. 

Upon  the  given  edge  AB  construct  the 
square  ABCDy 

and  upon  the  sides  of  this  square  con- 
C struct  the  squares  E  B,  FC,  G  D,  HA  _L  to 
the  plane  ABC  D. 

Then  A  G  is  the  regular  hexahedron  required. 

III.    To  construct  a  regular  octahedron. 

Upon  the  given  edge  A  B  construct 
the  square  ABC D. 

Through  its  centre  0  pass  a  J_  to 
its  plane  ABC D. 

In  this  _L  take  two  points  E  and  F, 
one  above  and  the  other  below  the  plane, 

so  that  A  E  and  A  F  are  each  equal 
toAB. 

Join  E  and  F  to  each  of  the  vertices  of  the  square. 

Then  E  ABC  D  F  is  the  regular  octahedron  required. 

For,  the  edges  are  all  equal, 

and  hence  the  faces  are  equal  equilateral  A. 

And,  since  the  A  D  EF  and  D  A  C  are  equal,  §  108 

D  EBF  is  a  square  and  the  pyramid  A-D  EB F  is  equal  in 
all  its  parts  to  the  pyramid  E-A  BCD. 

Hence,  the  polyhedral  A  A  and  E  are  equal. 

In  like  manner  all  the  polyhedral  A  of  the  figure  are  equal. 


E 


A 


D 


§  450 


324 


GEOMETRY. BOOK    VII. 


IV.    To  construct  a  regular  dodecahedron. 

Upon  A B  construct  the  regular  pentagon  ABODE.  §  395 

On  each  side  of  this  pentagon  construct  an  equal  pentagon, 
so  inclined  that  trihedral  A  shall  be  formed  at  A,  B,  0,  D,  E. 

The  convex  surface  thus  formed  is  composed  of  six  regular 
pentagons. 

In  like  manner,  upon  an  equal  pentagon  A'  B'  C  D'  E'  con- 
struct an  equal  convex  surface. 

Apply  one  of  these  surfaces  to  the  other,  with  their  convexi- 
ties turned  in  opposite  directions,  so  that  P'  0'  and  P'  Q1  shall 
fall  upon  P  0  and  P  Q. 

Then  every  face  Z  of  the  one  will,  with  two  consecutive 
face  A  of  the  other,  form  a  trihedral  Z. 

The  solid  thus  formed  is  the  regular  dodecahedron  required. 

For,  the  faces  are  all  regular  pentagons,  Cons, 

and  the  polyhedral  A  are  all  equal.  §  499 

D  D' 


G  G' 

V.    To  construct  a  regular  icosahedron. 

Upon  A  B  construct  the  regular  pentagon  ABODE.  §  395 
At  its  centre  0  erect  a  _U  to  its  plane. 
In  this  X  take  P  so  that  PA  =  A  B. 


REGULAR   POLYHEDRONS. 


325 


Join  P  with  each  of  the  vertices  of  the  pentagon  ; 

thus  forming  a  regular  pentagonal  pyramid  whose  vertex  is  P, 
and  whose  dihedral  A  formed  on  the  edges  PA,  P JB,  PC,  etc. 
are  all  equal.  §  571 

Taking  A  and  B  as  vertices,  construct  two  pyramids  each 
equal  to  the  first,  and  having  for  bases  BPEFGwAAGECP 
respectively. 

There  will  thus  be  formed  a  convex  surface  consisting  of  ten 
equal  equilateral  A. 

In  like  manner  upon  an  equal  pentagon  A'  B'  C  D1  E'  con- 
struct an  equal  convex  surface. 

Apply  one  of  these  surfaces  to  the  other  with  their  convexi- 
ties turned  in  opposite  directions,  so  that  every  combination  of 
two  face  A  of  the  one,  as  P'  D'  C,  P*  D'  E',  shall  with  a  combi- 
nation of  three  face  A  of  the  other,  as  BCH,  BCP,  PCD, 
form  a  pentahedral  Z. 

The  solid  thus  formed  is  the  regular  icosahedron  required. 

For,  the  faces  are  all  equal ;  Cons. 

and  the  polyhedral  A  are  all  equal,  §  571 

Q.  E.  D. 


TETRAHEDRON. 


HEX  AH 

EDRON. 

ICOSAHEDRON. 


DODECAHEDRON. 


593.  Scholium.  The  regular  polyhedrons  can  be  formed 
thus : 

Draw  the  above  diagrams  upon  card-board.  Cut  through 
the  exterior  lines  and  half  through  the  interior  lines.  The  fig- 
ures will  then  readily  bend  into  the  regular  forms  required. 


326  geometry. book  vii. 

Supplementary  Propositions. 
Proposition  XXV.     Theorem.    (Euler's.) 
594.   In  any  polyhedron  the  number  of  its  edges  in- 
creased by  two  is  equal  to  the  number  of  its  vertices  increased 
by  the  number  of  its  faces. 

Let  E  denote  the  number  of  edges  of  any  polyhedron; 
V  the  number  of  its  vertices,  F  the  number  of  its 
faces. 

We  are  to  prove  E  +  2  =  V  +  F. 

S  Beginning  with  one  face  ABODE, 

we  have  E  =  V. 

Annex  a  second  face  SAB  by  ap- 
plying one  of  its  edges  to  an  edge  of 
the  first  face. 

There  is  formed  a  surface  having 
one  edge  A  B,  and  two  vertices  A  and 
)D  B  common  to  both  faces. 

.*.  whatever  the  number  of  the 
B  C      sides  of  the  new  face,  the  whole  num- 

ber of  edges  is  now  one  more  than  the  whole  number  of  ver- 
tices. 

.-.for  2  faces  E=  V+  1. 

Annex  a  third  face,  SBC,  adjacent  to  each  of  the  former. 

The  new  surface  will  have  two  edges  SB  and  B  C, 

and  three  vertices  S,  B  and  C,  in  common  with  the  preced- 
ing surface. 

.*.  the  increase  in  the  number  of  edges  is  again  one  more 
than  the  increase  in  the  number  of  vertices. 

According  to  the  same  law,  for  an   incomplete  surface  of 
F—\  faces 

E=  V+  F-2. 

When  we  add  the  last  face  SEA,  necessary  to  complete  the 
surface, 

its  edges  SE,  SA  and  A  E,  and  its  vertices  S,  E  and  A 
will  be  in  common  with  the  preceding  surface. 

.*.  in  a  polyhedron  of  F  faces  E  —  V  +  F  —  2. 
,-.E+  2=  V+  F. 

Q.  E.  D. 


POLYHEDRONS.  327 


Proposition  XXVI.     Theorem. 

595.  The  sum  of  all  the  angles  of  the  faces  of  any  poly- 
hedron is  equal  to  four  right  angles  taken  as  many  times  as 
the  polyhedron  has  vertices  less  two. 

Let  E  denote  the  number  of  edges,  V  the  number  of 
vertices,  F  the  number  of  faces,  and  S  the  sum  of 
all  the  angles  of  the  faces  of  any  polyhedron. 

We  are  to  prove      S  =  4  rt.  A  X  ( V—  2). 

Since  E  denotes  the  number  of 
the  edges  of  the  polyhedron, 

2  E  will  denote  the  whole  num- 
ber of  sides  of  all  its  faces,  con- 
sidered as  sides  of  independent  poly- 
gons. 

A{ 

And  since  the  sum  of  all  the 
interior  and  exterior  A  of  each  poly-        B  c 

gon  is  equal  to  2  rt.  A  taken  as  many  times  as  it  has  sides, 

the  sum  of  the  interior  and  exterior  A  of  all  the  faces  is 
equal  to  2  rt.  A  X  2  E. 

And  since  the  sum  of  the  exterior  A  of  each  face  is 
4  rt.  A,  §  159 

the  sum  of  the  exterior  A  of  all  the  faces  is  equal  to 
4  rt.  A  X  F. 

.\  S+  4  rt  A  X  F*=*  2  it  A  X  2  E. 

That  is,     .         S  =  4  rt.  A  X  (E  —  F). 

Since  E  +  2  =  V  +  F,  §  594 

E-  F=  F-2, 

.'.  £=4rt.  A  X  (F-2).  q.  E.  d. 


328  geometry. book  vii. 

On  the  Cylinder. 

596.  Def.  A  Cylindrical  surface  is  a  curved  surface  gen- 
erated by  a  moving  straight  line  which  continually  touches  a 
given  curve  and  in  all  its  positions  is  parallel  to  a  given  fixed 
straight  line  not  in  the  plane  of  the  curve. 


Thus,  the  surface  ABC ' D,  generated  by  the  moving  line 
A  D  continually  touching  the  curve  ABC  and  always  parallel 
to  a  given  straight  line  M,  is  a  cylindrical  surface. 

597.  Def.  The  moving  line  is  called  the  Generatrix;  the 
curve  which  directs  the  motion  of  the  generatrix  is  called  the 
Directrix ;  the  generatrix  in  any  position  is  called  an  Element 
of  the  surface. 

The  generatrix  may  be  indefinite  in  extent,  and  the  direc- 
trix a  closed  or  an  open  curve.  In  elementary  geometry  the 
directrix  is  considered  a  circle. 

598.  Def.  A  Cylinder  is  a  solid  bounded  by  a  cylindrical 
surface  and  two  parallel  planes. 

599.  Def.    The  Bases  of  a  cylinder  are  its  plane  surfaces. 

600.  Def.  The  Lateral  surface  of  a  cylinder  is  its  cylindri- 
cal surface. 

601.  Def.  The  Axis  of  a  cylinder  is  the  straight  line  join- 
ing the  centres  of  its  bases. 


CYLINDERS.  329 


602.  Def.  The  Altitude  of  a  cylinder  is  the  perpendicular 
distance  between  the  planes  of  its  bases. 

603.  Def.  A  Section  of  a  cylinder  is  a  plane  figure  whose 
boundary  is  the  intersection  of  its  plane  with  the  surface  of  the 
cylinder. 

604.  Def.  A  Right  section  of  a  cylinder  is  a  section  per- 
pendicular to  the  elements. 

605.  Def.    A  Radius  of  a  cylinder  is  the  radius  of  the  base. 

606.  Def.  A  Right  cylinder  is  a  cylinder  whose  elements 
are  perpendicular  to  its  bases.  Any  element  of  a  right  cylinder 
is  equal  to  its  altitude. 

607.  Def.  An  Oblique  cylinder  is  a  cylinder  whose  elements 
are  oblique  to  its  bases.  Any  element  of  an  oblique  cylinder  is 
greater  than  its  altitude. 

608.  Def.  A  Cylinder  of  Revolution  is  a  cylinder  generated 
by  the  revolution  of  a  rectangle  about  one  side  as  an  axis. 

609.  Def.  Similar  cylinders  of  revolution  are  cylinders 
generated  by  similar  rectangles  revolving  about  homologous  sides. 

610.  Def.  A  Tangent  line  to  a  cylinder  is  a  straight  line 
which  touches  the  surface  of  the  cylinder,  but  does  not  intersect  it. 

611.  Def.  A  Tangent  plane  to  a  cylinder  is  a  plane  which 
embraces  an  element  of  the  cylinder  without  cutting  the  sur- 
face. The  element  embraced  by  the  tangent  plane  is  called 
the  Element  of  Contact. 

612.  Def.  A  prism  is  inscribed  in  a  cylinder  when  its 
lateral  edges  are  elements  of  the  cylinder  and  its  bases  are  in- 
scribed in  the  bases  of  the  cylinder. 

613.  Def.  A  prism  is  circumscribed  about  a  cylinder  when 
its  lateral  faces  are  tangent  to  the  cylinder  and  its  bases  are  cir- 
cumscribed about  the  bases  of  the  cylinder. 


330 


GEOMETRY. BOOK    VII. 


Proposition  XXVII.     Theorem. 

614.    Every  section  of  a  cylinder  made  by  a  plane  pass- 
ing through  an  element  is  a  parallelogram. 

G 


K?\ 


^^7 


Let  ABC  D  be  a  section  of  the  cylinder  A  G,  made  by 
a  plane  passing  through  A  D. 

We  are  to  prove  the  section  A  B  G D  a  parallelogram. 

The  line  B  G,  in  which  the  cutting  plane  intersects  the 
curved  surface  a  second  time,  is  an  element ; 

for,  if  through  the  point  B  a  line  be  drawn  II  to  A  D, 

it  will  be  an  element  of  the  surface. 

It  will  also  lie  in  the  plane  A  G. 

This  element,  lying  in  both  the  cylindrical  surface  and  plane 
surface,  is  their  intersection. 

Now  A  D  is  II  to  B  C, 

(being  elements  of  the  cylinder), 

and  A  B  is  II  to  D  G,  §  465 

(the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines). 

.-.  the  section  ABGD  is  a  O.  §125 

Q.  E.  D. 

615.  Corollary.    Every  section  of  a  right  cylinder  embrac- 
ing an  element  is  a  rectangle. 


CYLINDERS. 


331 


Proposition  XXVIII.     Theorem. 
616.    The  bases  of  a  cylinder  are  equal. 
£_ C 


Let  ABE  and  DGG  be  the  bases  of  the  cylinder  A  G. 
We  are  to  prove       A  B  E  =  D  C  G. 

Any  sections  A  G  and  A  G,  embracing  A  D,  an  element  of 

§614 

D  G  and  A  E  =  D  G. 

BG  is  II  to  EG, 
(each  being  II  to  AD). 

BC  =  EG, 

.'.  EG  is  a  O. 

.\EB  =  GG, 

\AEAB  =  A  GDG. 


the  cylinder,  are  UJ. 

.'.AB 
Now 


Also 


§134 
§459 

§464 
§  136 
§  134 
§  108 

Apply  the  upper  base  to  the  lower  base,  so  that  D  G  will 
coincide  with  A  B. 

Then  A  GDC  will  coincide  with  A  EAB,  and  point  G 
will  fall  upon  point  E. 

That  is,  any  point  G  in  the  perimeter  of  the  upper  base  will 
coincide  with  the  point  in  the  same  element  in  the  lower  base. 
.*.  the  bases  coincide,  and  are  equal. 

Q.  E.  D. 

617.  Corollary  1.  Any  two  parallel  sections  ABC  and 
A'  B'  C,  cutting  all  the  elements  of  a  cylinder  E  F,  are  equal. 
For  these  sections  are  the  bases  of  the  cylinder  A  C. 

618.  Cor.  2.  Any  section  of  a  cylinder  parallel  to  the  base 
is  equal  to  the  base. 


332  GEOMETRY. BOOK    VII. 

Proposition  XXIX.     Theorem. 
619.    The  lateral  area  of  a  cylinder  is  equal   to    the 
product  of  the  perimeter  of  a  right  section  of  the  cylinder  by 
an  element  of  the  surface.        ^ 


Let  ABC D  E  be  the  base,  and  A  A'  any  element  of  the 
cylinder  A  C ;  and  let  the  curve  abcdebe  any  right 
section  of  its  surface. 

Denote  the  perimeter  of  the  right  section  by  P, 
and  the  lateral  surface  of  the  cylinder  by  & 
We  are  to  prove         S  =  P  X  A  A'. 

Inscribe  in  the  cylinder  a  prism  whose  right  section  abcde 

will  be  a  polygon  inscribed  in  the  right  section  a  b  c  d  e  of  the 

cylinder.  §  604 

Denote  the  lateral  area  of  the  prism  by  s, 

and  the  perimeter  of  its  right  section  by  p. 

Then  s=pXAA',  §524 

{the  lateral  area  of  a  prism  is  equal  to  the  product  of  the  perimeter  of  a  right 
section  by  a  lateral  edge). 

Now  let  the  number  of  lateral  faces  of  the  inscribed  prism 
be  indefinitely  increased, 

the  new  edges  continually  bisecting  the  arcs  in  the  right 
section. 

Then  p  approaches  P  as  its  limit, 
and  s  approaches  S  as  its  limit. 
But,  however  great  the  number  of  faces, 
$=p  X  A  A1. 
.'.S=PX  AA\  §199 

Q.  E.  D. 


CYLINDERS. 


333 


II 


IV 


620.  Corollary  1.  The  lateral  area  of  a  right  cylinder  is 
equal  to  the  product  of  the  perimeter  of  its  base  by  its  altitude. 

621.  Cor.  2.  Let  a  cylinder  of  revolution  be  generated  by 
the  rectangle  whose  sides  are  R  and  H  revolving  about  the 
side  H. 

Then  R  is  the  radius  of  the  base  of  the  cylinder,  and  H  the 
altitude  of  the  cylinder. 

The  perimeter  of  the  base  is  2  n-  R ;  §  381 

hence,  S  =  2  it  R  X  H. 

The  area  of  each  base  is  tt  R2 ;  §  381 

hence,  the  total  area  T  of  a  cylinder  of  revolution  is  ex- 
pressed by 

T=2irRXH+27rR2  =  27rR(H+R). 

622.  Cor.  3.  Let  S,  S'  denote  the  lateral  areas  of  two  simi- 
lar cylinders  of  revolution ; 

T,  T'  their  total  areas ;  R,  R'  the  radii  of  their  bases ;  //,  H' 
their  altitudes. 

Since  the  generating  rectangles  are  similar,  we  have 

#=^=#jfljff  -266 

H'      R1      H'  +  R1  * 


S  _2irRII 
S'~2  7rR'H' 


x^  =  ^2 


R* 


H'      H'*      R'2 ' 


and  —-    2^(#+-ft)  _&  (H+  R\_  H2 _  R2 
T'~2it  R'  {H'  +  R')~R'  \H'  +  R1)  ~  E12  ~  R'2' 

That  is,  the  lateral  areas,  or  the  total  areas,  of  similar  cylin- 
ders of  revolution  are  to  each  other  as  the  squares  of  their  altitudes, 
or  as  the  squares  of  the  radii  of  their  bases. 


334 


GEOMETRY BOOK    VII. 


Proposition  XXX.     Theorem. 
623.    The  volume  of  a  cylinder  is  equal  to  the  product  of 
its  base  by  its  altitude. 


Let  V  denote  the  volume  of  the  cylinder  A  G,  B  its 
base,  and  H  its  altitude. 


We  are  to  prove 


V=BX  H. 


Let  V  denote  the  volume  of  the  inscribed  prism  A  G,  B'  its 
base,  and  H  will  be  its  altitude. 


Then 


V'  =  B'X  H, 


§543 


{the  volume  of  a  prism  is  equal  to  the  product  of  its  base  by  its  altitude). 
Now,  let  the  number  of  lateral  faces  of  the  inscribed  prism 
be  indefinitely  increased,  the  new  edges  continually  bisecting 
the  arcs  of  the  bases. 

Then  B'  approaches  B  as  its  limit, 

and  V  approaches  V  as  its  limit. 

But  however  great  the  number  of  the  lateral  faces, 

V'  =  B'X  H. 

.'.V=BXH.  §199 

Q.  E.  D. 


CYLINDERS.  335 

624.  Corollary  1.  Let  Vbe  the  volume  of  a  cylinder  of 
revolution,  R  the  radius  of  its  base,  and  H  its  altitude. 

Then  the  area  of  its  base  is  rr  R2,  §  381 

.'.  V=7rE2X  h. 

625.  Cor.  2.  Let  V  and  V  be  the  volumes  of  two  similar 
cylinders  of  revolution,  R  and  R'  the  radii  of  their  bases,  H  and 
H'  their  altitudes. 

Since  the  generating  rectangles  are  similar,  we  have 

B_=  ^. 
H'       R'} 

V  ttR2H       R2     H       H*      R3 
and       = =  —  v  —  = a . 

V  itR^H'     R'*     H'     H'8     R'* 

That  is,  the  volumes  of  similar  cylinders  of  revolution  are  to 
each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the 
radii  of  their  bases. 

Ex.  1.  Required,  the  entire  surface  and  volume  of  a  cylin- 
der of  revolution  whose  altitude  is  30  inches,  and  whose  base 
is  a  circle  of  which  the  diameter  is  20  inches. 

2.  Eequired,  the  volume  of  a  right  truncated  triangular 
prism  the  area  of  whose  base  is  40  inches,  and  whose  lateral 
edges  are  10,  12,  and  15  inches,  respectively. 

3.  Let  E  denote  an  edge  of  a  regular  tetrahedron ;  show 
that  the  altitude  of  the  tetrahedron  is  equal  to  E  y/~|~;  that  the 
surface  is  equal  to  E'2  ^~3 ;  and  that  the  volume  is  equal  to 

4.  Required,  the  number  of  quarts  that  a  cylinder  of  revo- 
lution will  contain  whose  height  is  20  inches,  and  whose  diame- 
ter is  12  inches. 

5.  Given  S,  the  surface  of  a  cube,  find  its  edge,  diagonal, 
and  volume.     What  do  these  become  when  S  =  54 1 


336 


GEOMETRY. BOOK    VII. 


Proposition  XXXI.     Problem. 

626.    Through  a  given  point  to  pass  a  plane 
given  c$ 


to  a 


Case  I.  —  When  the  given  point  is  in  the  curved  surface  of  the  cylinder. 

Let  AC  be  a  given  cylinder,  and  let  the  given  point 
be'  a  point  in  the  element  A  A'. 

It  is  required  to  pass  a  plane  tangent  to  the  cylinder  and  em- 
bracing the  element  A  A'. 

Draw  the  radius  0  A,  and  A  T  tangent  to  the  base ; 

and  pass  a  plane  R  T'  through  A  A'  and  A  T. 

The  plane  R  T1  is  the  plane  required. 

For,  through  any  point  P  in  this  plane,  not  in  the  ele- 
ment A  A', 

pass  a  plane   II  to  the  base,  intersecting  the  cylinder  in 
the  O  MN, 

and  the  plane  R  T>  in  MP. 

Prom  the  centre  of  the  O  M  N  draw  Q  M. 

MP  and  M  Q  are  II  respectively  to  A  T  and  A  0,         §  465 
(the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines)  ; 

r.ZPMQ  =  Z  TAO,  §  462 

(two  A  not  in  the  same  plane,  having  their  sides  II  and  lying  in  the  same 
direction,  are  equal). 


CYLINDERS.  337 

.'.  P  M  is  tangent  to  the  O  MN  at  M.  §  186 

.'.  P  lies  without  the  O  M N, 

and  hence  without  the  cylinder. 

.*•  the  plane  R  T'  does  not  cut  the  cylinder,  and  is  tangent 
to  it. 

Case  IT.  —  When  the  given  point  is  urithout  the  cylinder. 

Let  P  be  the  given  point. 

It  is  required  to  pass  a  plane  through  P  tangent  to  th% 
cylinder. 

Through   P  draw  the  line  P  T  II  to  the  elements  of  the 
cylinder, 

meeting  the  plane  of  the  base  at  T. 

From  T  draw  TA  and  TC  tangents  to  the  base.     §  240 

Through  P  T  and  the  tangent  TA  pass  a  plane  R  V. 

Since  A  A1  is  II  to  P  T,  Cons. 

the  plane  R  T',  passing  through  P  T  and  the  point  A  will 
contain  the  element  A  A', 

(two  II  lives  He  in  the  same  plane). 

And,  since  R  V  also  contains  the  tangent  A  T, 

it  is  a  tangent  plane  to  the  cylinder. 

In  like  manner,  the  plane  T  S',  passed  through  P  T  and  the 
tangent  line  T  C, 

is  a  tangent  plane  to  the  cylinder. 

Q.  E.  F. 

627.  Corollary  1.    The  intersection  of  two  tangent  planes 
to  a  cylinder  is  parallel  to  the  elements  of  the  cylinder. 

628.  Cor.  2.    Any  straight  line  drawn  in  a  tangent  plane, 
and  cutting  the  element  of  contact,  is  tangent  to  the  cylinder. 


338  geometry. book  vii. 

On  the  Cone. 

629.  Def.  A  Conical  surface  is  a  surface  generated  by  a 
moving  straight  line  continually  touching  a  given  curve  and 
passing  through  a  fixed  point  not  in  the  plane  of  the  curve. 

Thus  the  surface  generated  by  the  mov- 
ing line  A  A'  continually  touching  the  curve 
ABC D,  and  passing  through  the  fixed  point 
S,  is  a  conical  surface. 

630.  Def.  The  moving  line  is  called 
the  Generatrix ;  the  curve  which  directs  the 
motion  of  the  generatrix  is  called  the  Di- 
rectrix ;  the  generatrix,  in  any  position,  is 
called  an  Element  of  the  surface. 

631.  Def.  A  conical  surface  generated 
by  an  indefinite  straight  line  consists  of  two 
portions,  called  Nappes,  one  the  Lower,  the 
other  the  Upper  Nappe. 

632.  Def.  A  Cone  is  a  solid  bounded  by  a  conical  surface 
and  a  plane. 

633.  Def.  The  Lateral  surface  of  a  cone  is  its  conical  sur- 
face. 

634.  Def.    The  Base  of  a  cone  is  its  plane  surface. 

635.  Def.  The  Vertex  of  a  cone  is  the  fixed  point  through 
which  all  the  elements  pass. 

636.  Def.  The  Altitude  of  a  cone  is  the  perpendicular  dis- 
tance between  its  vertex  and  the  plane  of  its  base. 

637.  Def.  A  Section  of  a  cone  is  a  plane  figure  whose 
boundary  is  the  intersection  of  its  plane  with  the  surface  of  the 
cone. 

638.  Def.  A  Right  section  of  a  cone  is  a  section  perpen- 
dicular to  the  axis. 

639.  Def.    A  Circular  cone  is  a  cone  whose  base  is  a  circle. 

640.  Def.  The  Axis  of  a  cone  is  the  straight  line  joining 
its  vertex  and  the  centre  of  its  base. 

641.  Def.  A  Right  cone  is  a  cone  whose  axis  is  perpen- 
dicular to  its  base.  The  axis  of  a  right  cone  is  equal  to  its 
altitude. 

642.  Def.  An  Oblique  cone  is  a  cone  whose  axis  is 
oblique  to  its  base.  The  axis  of  an  oblique  cone  is  greater 
than  its  altitude. 


coxes.  339 


643.  Def.  A  Cone  of  Revolution  is  a  cone  generated  by  the 
revolution  of  a  right  triangle  about  one  of  its  perpendicular  sides 
as  an  axis. 

The  side  about  which  the  triangle  re- 
volves is  the  axis  of  the  cone  ;  the  other  per- 
pendicular generates  the  base,  the  hypotenuse 
generates  the  conical  surface.  Any  position 
'»i'  the  hypotenuse  is  an  element,  and  any 
element  is  called  the  slant  height. 

644.  Def.  Similar  cones  of  revolution 
are  cones  generated  by  the  revolution  of  simi- 
lar right  triangles  about  homologous  perpen- 
dicular sides. 

645.  Def.  A  Truncated  cone  is  the  portion  of  a  cone 
included  between  the  base  and  a  section  cutting  all  the  elements. 

646.  Def.  A  Frustum  of  a  cone  is  a  truncated  cone  in 
which  the  cutting  section  is  parallel  to  the  base. 

647.  Def.  The  base  of  the  cone  is  called  the  Lower  base  of 
the  frustum,  and  the  parallel  section  the  Upper  base. 

648.  Def.  The  Altitude  of  a  frustum  is  the  perpendicular 
distance  between  the  planes  of  its  bases. 

649.  Def.  The  Lateral  surface  of  a  frustum  is  the  portion 
of  the  lateral  surface  of  the  cone  included  between  the  bases  of 
the  frustum. 

650.  Def.  The  Slant  height  of  a  frustum  of  a  cone  of  revo- 
lution is  the  portion  of  any  element  of  the  cone  included  between 
the  bases. 

651.  Def.  A  Tangent  line  to  a  cone  is  a  line  having  only 
one  point  in  common  with  the  surface. 

652.  Def.  A  Tangent  plane  to  a  cone  is  a  plane  embracing 
an  element  of  the  cone  without  cutting  the  surface.  The  element 
embraced  by  the  tangent  plane  is  called  the  Element  of  Contact. 

653.  Def.  A  pyramid  is  inscribed  in  a  cone  when  its  lat- 
eral edges  are  elements  of  the  cone  and  its  base  is  inscribed  in 
the  base  of  the  cone. 

654.  Def.  A  pyramid  is  circumscribed  about  a  cone  when 
its  lateral  faces  are  tangent  to  the  cone  and  its  base  is  circum- 
scribed about  the  base  of  the  cone. 


340 


GEOMETRY. BOOK    VII. 


Proposition  XXXII.     Theorem. 
655.    Every  section  of  a  cone  made  by  a  plane  passing 

through  its  vertex  is  a  triangle. 

S 


Let  SBD   be  a  section  of  the  cone  S-ABG  through 
the  vertex  S. 

We  are  to  prove  the  section  SB  D  a  triangle. 

The  straight  lines  joining  S  with  B  and  D  are  elements  of 
the  surface.  §  630 

They  also  lie  in  the  cutting  plane, 
(for  their  extremities  lie  in  tJie  plane). 

Hence,  they  are  the  intersections  of  the  conical  surface  with 
the  plane  of  the  section. 

B  D  is  also  a  straight  line,  §  446 

(the  intersection  of  two  planes  is  a  straight  line). 

.'.  the  section  SBD  is  a  A. 


Q.  E.  D 


coxes.  341 


Proposition  XXXIII.     Theorem. 

656.    Every  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  is  a  circle. 


Let  the  section  a  b  c  of  the  circular  cone  S-A  B  C  be 
parallel  to  the  base. 

We  are  to  prove  that  a  b  c  is  a  circle. 

Let  0  be  the  centre  of  the  base,  and  let  o  be  the  point  in 
which  the  axis  S  0  pierces  the  plane  of  the  II  section. 

Through  SO  and  any  number  of  elements,  SA,  SB,  etc., 
pass  planes  cutting  the  base  in  the  radii  0  A,  OB,  etc., 

and  the  section  a  b  c  in  the  straight  lines  o  a,  ob,  etc. 

Now  o  a  and  o  b  are  II  respectively  to  0  A  and  0  B,   §  465 
{the  intersections  of  two  II  planes  by  a  third  plane  are  II  lines). 
/.the  A   So  a  and   Sob  are  similar  respectively  to  the 
ASOAandSOB,  .  §279 

and  their  homologous  sides  give  the  proportion 


oa       / 'S '  o\       ob 
OA  =  \S0)  =  OB  ' 


But  OA  =  OB;  §163 

.".  o  a  =  o  b. 
That  is,  all  the  straight  lines  drawn  from  o  to  the  perimeter 
of  the  section  are  equal. 

.'.  the  section  a  b  c  is  a  O. 

Q.  E.  D. 

657.  Corollary.    The  axis  of  a  circular  cone  passes  through 
the  centres  of  all  the  sections  which  are  parallel  to  the  base. 


842 


GEOMETRY. 


BOOK  vn. 


Proposition  XXXIV.     Theorem. 

658.  The  lateral  area  of  a  cone  of  revolution  is  equal  to 
one-half  the  product  of  the  circumference  of  its  base  by  the 
slant  height. 


Let  A-E F G  H K  be  a  cone  generated  by  the  revolution 
of  the  right  triangle  A  OE  about  AO  as  an  axis,  and 
let  S  denote  its  lateral  area,  G  the  circumference 
of  its  base   and  L  its  slant  height. 

We  are  to  prove  S  =  \  G  X  L. 

Inscribe  on  the  base  any  regular  polygon  E FG  H K, 

and  upon  this  polygon  as  a  base  construct  the  regular  pyra- 
mid A-E  F  G  UK  inscribed  in  the  cone. 

Denote  the  lateral  area  of  this  pyramid  by  s,  the  perimeter 
of  its  base  by  p,  its  slant  height  by  I, 


Then 


s  =  \p  X  I, 


569 


{the  lateral  area  of  a  regular  pyramid  is  equal  to  one-half  the  product  of  the 
perimeter  of  its  base  by  the  slant  height). 

Now,  let  the  number  of  the  lateral  faces  of  the  inscribed 
pyramid  be  indefinitely  increased, 


cones.  343 


the  new  edges  continually  bisecting  the  arcs  of  the  base. 

Then  p,  s  and  I  approach  C,  8  and  L  respectively  as  their 
limits. 

But  however  great  the  number  of  lateral  faces  of  the 
pyramid, 

s  =  \p  X  I. 

>\J3m*iOxL.  §199 

Q.  E.  D. 


659.  Corollary  1.  If  R  be  the  radius  of  the  base,  we 
have  C=27ri?(§381).  Therefore  S=±(2 vR  X  L)  =  vRL. 
Also,  since  the  area  of  the  base  is  ttR7,  the  total  area  Tof  the 
cone  is  expressed  by 

T=  ttRL  +  ttR2  =  irR(L  +  R). 

660.  Cor.  2.  Let  S  and  S1  denote  the  lateral  areas  of 
two  Bimilar  cones  of  revolution,  T  and  T  their  total  areas, 
R  and  R'  the  radii  of  their  bases,  iiT and  W  their  altitudes, 
L  and  V  their  slant  heights.  Since  the  generating  triangles 
are  Bimilar,  we  have 


L      H       R      R  +  L 


L>      H>      R'      W  +  L'  ' 


irRL  _R       L       L2       R*       H* 

~~~  ~z~.  X  — r~. ' 


206 


£'      ttR'L'      R'      L'      L'2       R'*     Hri' 

T  n  wRX(L+R)  _R       L  +  R  _  Z2  _  R*  _  IP 
T1     ir  R'  X  {L1  +  R')      R'     L'  +  R'      T*      R*      IF2' 

That  is  :  the  lateral  areas,  or  total  areas,  of  similar  cones  of 
revolution  are  to  each  other  as  the  squares  of  their  slant  heights,  the 
squares  of  their  altitudes,  or  the  squares  of  the  radii  of  their  bases. 


3U 


GEOMETRY. 


BOOK    VII. 


Proposition  XXXV.     Theorem. 

661.  The  lateral  area  of  the  frustum  of  a  cone  of  revo- 
lution is  equal  to  one-half  the  sum  of  the  circumferences  of  its 
buses  multiplied  by  the  slant  height. 


het  II BC-E  FG  be  the  frustum  of  a  cone  of  revolution, 
and  let  S  denote  its  lateral  area,  G  and  c  the  cir- 
cumferences of  its  lower  and  upper  bases,  R  and 
r  the  radii  of  the  bases,  and  L  the  slant  height. 

We  are  to  prove      S  =  \  (G  +  c)  X  L. 

Inscribe  in  the  frustum  of  the  cone  the  frustum  of  the  reg- 
ular pyramid  HBC-EFG, 

and  denote  the  lateral  area  of  this  frustum  by  s,  the  peri- 
meters of  its  lower  and  upper  bases  by  P  and  p  respectively,  and 
its  slant  height  by  /. 

Then  s  =  \  (P  +  p)  I,  §  570 

{the  lateral  area  of  the  frustum  of  a  regular  pyramid  is  equal  to  one-half 
the  sum  of  the  perimeters  of  its  bases  multiplied  by  the  slant  height). 

Now,  let  the  number  of  lateral  faces  be  indefinitely  in- 
creased, the  new  elements  constantly  bisecting  the  arcs  of  the 
bases. 


CONES. 


345 


Then  P,  p,  and  /,   approach  C,  c,  and  L,  respectively  as 
their  limits. 

But,  however  great  the  number  of  lateral  faces  of  the  frus- 
tum of  the  pyramid, 

•  -  J  (#  +  p)  X  I 


X  =  b(C+  c)X  L. 


§  199 

Q.  E.  D. 


662.  Corollary.  The  lateral  area  of  a  frustum  of  a  cone 
of  revolution  is  equal  to  the  circumference  of  a  section  equidistant 
from  its  bases  multiplied  by  its  slant  height. 

For  the  section  of  the  frustum  equidistant  from  its  bases 
cuts  the  frustum  of  the  regular  inscribed  pyramid  equidistant 
from  its  bases. 

Therefore  the  perimeter  /  LK  =  J  the  sum  of  the  perim- 
eters II  B  C  and  EFG.  §  142 

And  this  will  always  be  true,  however  great  the  number  of 
the  lateral  faces  of  the  frustum  of  the  pyramid. 

Hence,  circumference  ILK  =  J  the  sum  of  the  circumfer- 
ences HB  G  and  EFG.  §  199 


346 


GEOMETRY.  —  BOOK    VII. 


Proposition  XXXVI.     Theorem. 

G63.  Any  section  of  a  cone  parallel  to  the  base  is  io  the 
base  as  the  square  of  the  altitude  of  the  part  above  the  section 
is  to  the  square  of  the  altitude  of  the  cone. 


Let  B  denote  the  base  of  the  cone,  H  its  altitude, 
b  a  section  of  the  cone  parallel  to  the  base,  and 
h  the  altitude  of  the  cone  above  the  section. 

We  are  to  prove       B  :  b  :  :  IT2 :  Jr. 

Let  J51  denote  the  base  of  an  inscribed  pyramid,  b1  the  base 
of  the  pyramid  formed  in  the  section  of  the  cone. 

Then  B'  :  V  : :  IP  :  k\  §  566 

(any  section  of  a  pyramid  II  to  its  base  is  to  the  base  as  tJie  square  of  the  JL 
from  the  vertex  to  tlie  plane  of  the  section  is  to  tlie  square  of  tJie  altitude 
of  the  pyramid). 

Now  let  the  number  of  lateral  faces  of  the  inscribed  pyiv 
mid  be  indefinitely  increased, 

the  new  edges  continually  bisecting  the  arcs  in  the  base  of 
the  cone. 

Then  B'  and  b'  approach  B  and  b  respectively  as  their 
limits. 

But  however  great  the  number  of  lateral  faces  of  the  pyra- 
mid, 


B'  :b'  ::H2:  h\ 

.B:b  ::H*:h*, 

§199 

a  e.  d. 

CONES.  347 


Proposition  XXXVII.     Theorem. 
664.    The  volume  of  any  cone  is  equal  to  the  product  of 
one-third  of  its  base  by  its  altitude. 


Let  V  denote  the  volume,  B  the  base,  and  H  the  al- 
titude of  the  cone. 

We  are  to  prove        V  =  \  B  X  II. 

Let  the  volume  of  an  inscribed  pyramid  AC DEFG  be 
denoted  by  F,  and  its  base  by  B'. 

II  will  also  be  the  altitude  of  this  pyramid. 

Then  V'  =  %B'XH,  §574 

Now,  let  the  number  of  lateral  faces  of  the  inscribed  pyra- 
mid be  indefinitely  increased,  the  new  edges  continually  bisect- 
ing the  arcs  in  the  base  of  the  cone. 

Then  V  approaches  to  V  as  its  limit,  and  B'  to  B  as  its  limit. 

But  however  great  the  number  of  lateral  faces  of  the  pyramid, 


p-j 

B'  X  H. 

.  r— \ 

B  X  H. 

§  199 
Q.  E.  D. 

If  the 

cone  be  a  cone 

of  ] 

'evolution, 

of  the 

base,   then  B  = 

irtf 

1  (§381); 

665.  Corollary  1. 
and   R  be   the   radius 
.-.  V  =  \ttR2X  H. 

666.  Cor.  2.  Similar  cones  of  revolution  are  to  each  other 
as  the  cubes  of  their  altitudes,  or  as  the  cubes  of  the  radii  of  their 
bases.  For,  let  R  and  R'  be  the  radii  of  two  similar  cones 
of  revolution,  H  and  II'  their  altitudes,  V  and  V  their  volumes. 
Since  the  generating  triangles  are  similar,  we  have 

Hill'  ::  R  :  R'. 

V'~~hv  K/2X  H'~  Rh      H'~~  H'*~~  R'z' 


348  GEOMETRY. —  BOOK   VII. 

Proposition  XXXVIII.     Theorem. 
667.    A  frustum  of  any  cone  is  equivalent  to  the  sum  of 
three  cones  whose  common  altitude  is  the  altitude  of  the  frus- 
tum and  whose  bases  are  the  lower  base,  the  upper  base,  and  a 
mean  proportional  between  the  bases  of  the  frustum. 

h     j 


Let  V  denote  the  volume  of  the  frustum,  B  its  lower 
base,  b  its  upper  base,  and  H  its  altitude. 

We  are  to  prove     V=±H(B+b  +  y/  B  X  b). 

Let  V  denote  the  volume  of  an  inscribed  frustum  of  a  pyra- 
mid, B'  its  lower  base,  b'  its  upper  base. 
Its  altitude  will  also  be  H. 

Then,  V  =  J  H  {B'  +  b'  +  y/  B>  X  b'\  §  578 

(a  frustum  of  any  pyramid  is  ^  to  the  sum  of  three  pyramids  whose  common 
altitude  is  the  altitude  of  tfie  frustum,  and  whose  bases  are  the  lower 
base,  the  upper  base,  and  a  mean  proportional  between  the  bases  of  the 
frustum). 

Now,  let  the  number  of  lateral  faces  of  the  inscribed  frus- 
tum be  indefinitely  increased, 

the  new  edges  continually  bisecting  the  arcs  in  the  bases  of 
the  frustum  of  the  cone. 

But  however  great  the  number  of  lateral  faces  of  the  frus- 
tum of  the  pyramid, 

V  =  \H(B'  +  V  +  y/  B'  X  V. 

A  V=$H(B+  b+  s]  BXb).  §  199 

Q.  E.  D. 

668.  Corollary.  If  the  frustum  be  that  of  a  cone  of  revo- 
lution, and  R  and  r  be  the  radii  of  its  bases,  we  have  B  =  it  R2, 
and  b  =  n  r2, 

and  \/  BXb  =  7rJRr. 

.'.  V=  J  *•#(/?»+  r2+  Rr). 


BOOK   VIII. 


THE   SPHERE. 


On  Sections  and  TANGENTa 


669.  Def.  A  Sphere  is  a  solid  bounded  by  a  surface  all 
points  of  which  are  equally  distant  from  a  point  within  called 
the  centre.  A  sphere  may  be  generated  by  the  revolution  of  a 
semicircle  about  its  diameter  as  an  axis. 

670.  Def.  A  Radius  of  a  sphere  is 
the  distance  from  its  centre  to  any  point 
in  the  surface.  All  the  radii  of  a  sphere 
are  equal. 

671.  Def.  A  Diameter  of  a  sphere 
is  any  straight  line  passing  through  the 
centre  and  having  its  extremities  in  the 
surface  of  the  sphere.  All  the  diameters 
of  a  sphere  are  equal,  since  each  is  equal  to  twice  the  radius. 

672.  Def.  A  Section  of  a  sphere  is  a  plane  figure  whose  boun- 
dary is  the  intersection  of  its  plane  with  the  surface  of  the  sphere. 

673.  Def.  A  line  or  plane  is  Tangent  to  a  sphere  when  it  has 
one,  and  only  one,  point  in  common  with  the  surface  of  the  sphere. 

674.  Def.  Two  spheres  are  tangent  to  each  other  when  their 
surfaces  have  one,  and  only  one,  point  in  common. 

675.  Def.  A  polyhedron  is  circumscribed  about  a  sphere 
when  all  of  its  faces  are  tangent  to  the  sphere.  In  this  case  the 
sphere  is  inscribed  in  the  polyhedron. 

676.  Def.  A  2Joli/hedron  is  inscribed  in  a  sphere  when  all 
of  its  vertices  are  in  the  surface  of  the  sphere.  In  this  case  th  .■ 
sphere  is  circumscribed  about  the  polyhedron. 

677.  Def.  A  Cylinder  or  cone  is  circumscribed  about  a 
sphere  when  its  bases  and  cylindrical  surface,  or  its  base  and 
conical  surface,  are  tangent  to  the  sphere.  In  this  case  the 
sphere  is  inscribed  in  the  cylinder  or  cone. 


350 


GEOMETRY. 


BOOK   VIII. 


Proposition  I.     Theorem. 
678.  Every  section  of  a  sphere  made  by  a  plane  is  a  circle. 


Let  the  section  ABC  be  a,  plane  section  of  a  sphere 
whose  centre  is  0. 

We  are  to  prove     section  ABC  a  circle. 

From  the  centre  0  draw  0  D  _L  to  the  section,  and  draw 
the  radii  0  A,  OB,  0  C,  to  different  points  in  the  boundary  of 
the  section. 

In  the  rt.  A  0  D  A,  0  D  B  and  ODC, 

0  D  is  common,  and  0  A,  0  B  and  0  C  are  equal, 

(being  radii  of  the  sphere). 

.".  the  rt.  AODA,ODB  and  ODC  are  equal,  §  109 

(two  rt.  A  are  equal  when  they  have  a  side  and  hypotenuse  of  the  one  equal 
respectively  to  a  side  and  hypotenuse  of  the  other). 

.'.  DA,  D  B  and  D  C  are  equal, 
(being  homologous  sides  of  equal  &). 

.*.  the  section  A  B  C  is  a  circle  whose  centre  is  D. 

Q.  E.  D. 

679.  Corollary  1.  The  line  joining  the  centres  of  a  sphere 
and  a  circle  of  a  sphere  is  perpendicular  to  the  circle. 

680.  Cor.  II.  If  K,  r  and  p,  respectively,  denote  the 
radius  of  a  sphere,  the  radius  of  a  circle  of  a  sphere,  and  the  per- 
pendicular from  the  centre  of  the  sphere  to  the  circle,  then 
r  =  y  R2  —  p*.  Therefore  all  circles  of  a  sphere  equally  distant 
from  the  centre  are  equal,  and  of  two  circles  unequally  distant 
from  the  centre  of  the  sphere  the  more  remote  is  the  smaller. 

Again,  if  p  —  0,  then  r  =  R,  and  the  centre  of  the  sphere  and 
the  centre  of  the  circle  coincide ;  such  a  section  is  the  greatest 
possible  circle  of  the  sphere. 


THE    SPHERE. 


351 


G81.  Def.  A  Great  circle  of  a  sphere  is  a  section  of  the 
sphere  made  by  a  plane  passing  through  the  centre. 

682.  Def.  A  Small  circle  of  a  sphere  is  a  section  of  the 
sphere  made  by  a  plane  not  passing  through  the  centre. 

683.  Def.  An  Axis  of  a  circle  of  a  sphere  is  the  diameter 
(if  the  sphere  perpendicular  to  the  circle;  and  the  extremities  of 
the  axis  are  the  Poles  of  the  circle. 

684.  Every  great  circle  bisects  the  sphere.  For,  if  the  parts 
be  separated  and  placed  with  their  plane  sections  in  coinci- 
dence and  their  convexities  turned  the  same  way,  their  convex 
surfaces  will  coincide ;  otherwise  there  would  be  points  in  the 
spherical  surface  unequally  distant  from  the  centre. 

685.  Any  two  great  circles,  ABC D 
and  AECF,  bisect  each  other.  For  the 
intersection  A  C  of  their  planes  passes 
through  the  centre  of  the  sphere,  and  is 
a  diameter  of  each  circle. 


686.  An  arc  of  a  great  circle  may 
be  drawn  through  any  two  given  points 
A  and  E  in  the  surface  of  a  sphere.  For 
the  two  points  A  and  E,  and  the  centre 
0,  determine  the  plane  of  a  great  circle  whose  circumference 
passes  through  A  and  E.  §443 

If,  however,  the  two  given  points  are  the  extremities  A  and 
C  of  the  diameter  of  the  sphere,  the  position  of  the  circle  is  not 
determined.  For,  the  points  A,  O  and  C,  being  in  the  same 
straight  line,   an  infinite  number  of  planes  can  pass  through 


them. 


§441 


687.  One  circle,  and  only  one,  may  be  drawn  through  any 
three  given  points  on  the  surface  of  a  sphere.  For,  the  three 
points  determine  the  plane  which  cuts  the  sphere  in  a  circle. 


352 


GEOMETRY. BOOK    VIIT. 


Proposition  II.     Theorem. 

688.    A  plane  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  sphere. 


Let  0  be  the  centre  of  a  sphere,  and  M N  a  plane  per- 
pendicular to  the  radius  0  P,  at  its  extremity  P. 

We  are  to  prove     M  N  tangent  to  the  sphere. 

From  0  draw  any  other  straight  line  0  A  to  the  plane  M N. 

OP<OA,  §448 

(a  ±  is  the  shortest  distance  from  a  point  to  a  plane). 

.*.  point  A  is  without  the  sphere. 

But  0  A  is  any  other  line  than  0  P, 

.'.  every  point  in  the  plane  M N  is  without  the  sphere, 
except  P. 

.'.  M N  is  tangent  to  the  sphere  at  P.  §  673 

Q.  E.  D. 

689.  Corollary  1.  A  plane  tangent  to  a  sphere  is  perpen- 
dicular to  the  radius  drawn  to  the  point  of  contact. 

690.  Cor.  2.  A  straight  line  tangent  to  a  circle  of  a  sphere 
lies  in  a  plane  tangent  to  the  sphere  at  the  point  of  contact. 

691.  Cor.  3.  Any  straight  line  in  a  tangent  plane  through 
the  point  of  contact  is  tangent  to  the  sphere  at  that  point. 

692.  Cor.  4.  The  plane  of  any  two  straight  lines  tangent 
to  the  sphere  at  the  same  point  is  tangent  to  the  sphere  at  that 
point. 


THE    SPHERE.  353 


Proposition  III.     Problem. 
693.   Given  a  material  sphere  to  find  its  diameter. 

P 


P 


C> ^. 


<r 


d\ --■>&      /f...% 


\     \     /     / 

A'        ' 
pi  P' 

Let  P B P'  G  represent  a  material  sphere. 
It  is  required  to  find  its  diameter. 

From  any  point  P  of  the  given  surface,  with  any  opening  of 
the  compasses,  describe  the  circumference  A  B  G  on  the  surface. 

Then  the  straight  line  P B,  being  the  opening  of  the 
compasses,  is  a  known  line. 

Take  any  three  points  A,  B  and  G  in  this  circumference, 
and  with  the  compasses  measure  the  rectilinear  distances  A  B, 
B Guild  GA. 

Construct  the  A  A'  B'  C,  with  its  sides  equal  respectively 
to  A  B,  B  G  and  G  A.  §232 

Circumscribe  a  circle  about  the  A  A'  B'  G'.  §  239 

The  radius  D' B'  of  this  O  is  equal  to  the  radius  of  O  A  BG. 

Construct  the  rt.  Abdp,  having  the  hypotenuse  b  p=B  P, 
and  one  side  b  d  =  B'  D'. 

Draw  b  p1  J_  to  b  p,  and  meeting  p  d  produced  in  p'. 

Then  p  p'  is  equal  to  the  diameter  of  the  given  sphere. 

For,  if  we  bisect  the  sphere  through  P  and  B,  and  in  the 
section  draw  the  diameter  P  P'  and  chord  BP',  the  A  bpp', 
when  applied  to  A  B  P  P',  will  coincide  with  it. 

Q.  E.  F. 


354 


GEOMETRY.  —  BOOK    VIII. 


Proposition  IV.     Theorem. 

694.    Through  any  four  points  not  in  the  same  plane, 
one  spherical  surface  can  be  made  to  pass,  and  but  one. 


Let  A,  B,  C,  D,  be  four  points  not  in  the  same  plane. 

We  are  to  prove  that  one,  and  only  one,  spherical  surface  can 
be  made  to  pass  through  A,  B,  C,  D. 

Construct  the  tetrahedron  A  BC D,  having  for  its  vertices 
A,  B,  C,  D. 

Let  E  be  the  centre  of  the  circle  circumscribed  about  the 
face  ABC. 

Draw  EM  J_  to  this  face. 

Every  point  in  E  M  is  equally  distant  from  the  points  A, 
B  and  C,  §  450 

(oblique  lines  drawn  from  a  point  to  a  plane  at  equal  distances  from  tlie  foot 
of  the  _L  are  equal). 

Also,  let  F  be  the  centre  of  the  circle  circumscribed  about 
the  face  BC D ; 

and  draw  F  K  _L  to  this  face. 

Let  H  be  the  middle  point  of  B  C. 

Draw  EH  and  F H. 

Then  E  H  and  FH  are  JL  to  B  C  §  184 


THE    SPHERE.  355 


.'.the  plane  passed  through  EH  and  FH  is  1_  to  BC,  §  449 
(if  a  straight  line  be  JL  to  two  straight  lines  drawn  through  its  foot  in  a 
plane,  it  is  A.  to  the  plane,  and  in  this  case  the  plane  is  _L  to  the  line). 

Hence,  this  plane  is  also  _L  to  each  of  the  faces  ABC 
and  BOB,  §471 

(if  a  straight  line  be  ±  to  a  plane,  every  plane  ixisscd  through  that  line 
is  _L  to  the  plane). 

.'.  the  J«  E  M  and  FK  lie  in  the  plane  EHF. 

Hence  they  must  meet  unless  they  be  parallel. 

But  if  they  were  II,  the  planes  BCD  and  ABC  would  be 
one  and  the  same  plane,  which  is  contrary  to  the  hypothesis. 

Now  0,  the  point  of  intersection  of  the  J*  E  M  and  FK, 
is  equally  distant  from  A,  B  and  C ;  and  also  equally  distant 
from  B,  C  and  D  ; 

.*.  it  is  equally  distant  from  A,  B,  C  and  D. 

Hence,  a  spherical  surface,  whose  centre  is  0,  and  radius 
0  A,  will  pass  through  the  four  given  points. 

Only  one  spherical  surface  can  be  made  to  pass  through  the 
points  A,  B,  C  and  D. 

For  the  centre  of  such  a  spherical  surface  must  lie  in  both 
the  J*E M  and  FK. 

And,  since  0  is  the  only  point  common  to  these  J»,  0  is 
the  centre  of  the  only  spherical  surface  passing  through  A,  B,  C 
and  D. 

Q.  E.  D. 

695.  Corollary  1.  The  four  perpendiculars  erected  at  the 
centres  of  the  faces  of  a  tetrahedron  meet  at  the  same  point. 

696.  Cor.  2.  The  six  planes  perpendicular  to  the  six  edges 
of  a  tetrahedron  at  their  middle  point  will  intersect  at  the  same 
point. 


356  GEOMETRY.  —  BOOK   VIII. 

Proposition  V.     Theorem. 
697.  A  sphere  may  be  inscribed  in  any  given  tetrahedron. 
D 


B 
Let  ABCB  be  the  given  tetrahedron. 

We  are  to  prove  that  a  sphere  may  be  inscribed  in  ABC  B. 

Bisect  the  dihedral  A  at  the  edges  A  B,  B  C  and  A  C  by 
the  planes  0  A  B,  0  B  C  and  0  AC  respectively. 

Every  point  in  the  plane  0  A  B  is  squally  distant  from  the 
faces  ABC  and  ABB,  §  477 

For  a  like  reason,  every  point  in  the  plane  0  B  C  is  equally 
distant  from  the  faces  ABC  and  BBC; 

and  every  point  in  the  plane  0  A  C  is  equally  distant  from 
the  faces  A  B  C  and  A  B  C. 

.*.  0,  the  common  intersection  of  these  three  planes,  is 
equally  distant  from  the  four  faces  of  the  tetrahedron. 

.*.  a  sphere  described  with  0  as  a  centre,  and  with  the 
radius  equal  to  the  distance  of  0  to  any  face,  will  be  tangent  to 
each  face,  and  will  be  inscribed  in  the  tetrahedron.  §  673 

Q.  E.  D. 

698.  Corollary.  The  six  planes  which  bisect  the  six  dihe- 
dral angles  of  a  tetrahedron  intersect  in  the  same  point. 


On  Distances  Measured  on  the  Surface  of  the  Sphere. 

699.  Def.  The  distance  between  two  points  on  the  surface 
of  a  sphere  is  understood  to  be  the  arc  of  a  great  circle  joining 
the  points,  unless  otherwise  stated. 

700.  Def.  The  distance  from  the  pole  of  a  circle  to  any 
point  in  the  circumference  of  the  circle  is  the  Polar  distance  *>f 
the  circle. 


THE    SPHERE. 


357 


Proposition  VI.     Theorem. 

701.  The  distances  measured  on  the  surface  of  a  sphere 
from  all  points  in  the  circumference  of  a  circle  to  its  pole  are 
equal.  p 


A' 


A^lr 

/       ;i**^  n 

/      \S*»»«-_  V 

K^jp 

JQL_/    \ 

\                    *  1   '< 

if         j 

Let  P,F  be  the  poles  of  the  circle  ABC. 

We  are  to  prove  arcs  PA,  PB,  PC  equal. 

The  straight  lines  PA,  PB  and  PC  are  equal,     §  450 
{oblique  lines  drawn  from  a  point  to  a  plane  at  equal  distances  from  the  foot 
of  the  ±  are  equal) ; 

."•  the  arcs  P  A,  P  B  and  P  C  are  equal,  §  182 

(in  equal  (D  equal  chords  subtend  equal  arcs). 

In  like  manner  arcs  Pf  A,  P'  B  and  P'  C  are  equal. 

Q.  E.  D. 


702.  Corollary  1.  The  polar  distance  of  a  great  circle  is  a 
quadrant.  Thus,  arcs  PA',  PB',  P' A',  P' B',  polar  distances  of 
the  great  circle  A'  B'  C  D',  are  quadrants  ;  for  they  are  the  meas- 
ures of  the  right  angles  A'  OP,  B' 0 P,  A'  0 P',  B'  0 P',  whose 
vertices  are  at  the  centres  of  the  great  circles  PA'P'C,  PB'P'B'. 

703.  Scholium.  Every  point  in  the  circumference  of  a  small 
circle  is  at  unequal  distances  from  the  two  poles  of  the  circle. 


358 


GEOMETRY. 


BOOK    VIII. 


Proposition  VII.     Problem. 

704.    To  pass  a  circumference  of  a  great  circle  through 
any  two  points  on  the  surface  of  a  sphere. 


Let  A  and  B  be  any  two  points  on  the  surface  of  a 

sphere. 

It  is  required  to  pass  a  circumference  of  a  great  circle  through 
A  and  B. 

Prom  iasa  pole,  with  an  arc  equal  to  a  quadrant,  strike 
an  arc  a  b, 

and  from  B  as  a  pole,  with  the  same  radius,  describe  an  arc 
c  d,  intersecting  a  b  at  P. 

Then  a  circumference  described  with  a  quadrant  arc,  with 
P  as  a  pole,  will  pass  through  A  and  B  and  be  the  circumference 
of  a  great  circle. 

Q.  E.  F. 


705.  Corollary.  Through  any  two  points  on  the  surface 
of  a  sphere,  not  at  the  extremities  of  the  same  diameter,  only 
one  circumference  of  a  great  circle  can  be  made  to  pass. 

706.  Scholium.  By  means  of  poles  arcs  of  circles  may  be 
drawn  on  the  surface  of  a  sphere  with  the  same  facility  as  upon 
a  plane  surface,  and,  in  general,  the  methods  of  construction  in 
Spherical  Geometry  are  similar  to  those  of  Plane  Geometry. 
Thus  we  may  draw  an  arc  perpendicular  to  a  given  spherical  arc, 
bisect  a  given  spherical  angle  or  arc,  make  a  spherical  angle  equal 
to  a  given  spherical  angle,  etc.,  in  the  same  way  that  we  make 
analogous  constructions  in  Plane  Geometry. 


THE    SPHERE.  359 


Proposition  VIII.     Theorem. 

707.  The  shortest  distance  on  the  surface  of  a  sphere 
between  any  two  points  on  that  surface  is  the  arc,  not  greater 
than  a  semi-circumference,  of  the  great  circle  which  joins 
them. 


Let  A  B  be  the  arc  of  a  great  circle  which  joins  any 
two  points  A  and  B  on  the  surface  of  a  sphere ; 
and  let  A  C PQB  be  any  other  line  on  the  surface 
between  A  and  B. 

We  are  to  prove  arc  ABKACPQB. 

Let  P  be  any  point  in  A  C  P  Q  B. 

Pass  arcs  of  great  circles  through  A  and  P,  and  P 
and  B.  §  704 

Join  A,  P  and  B  with  the  centre  of  the  sphere  0. 
The  A  A  OB,  AOP  and  POB  are  the  face  A  of  the  tri- 
hedral A  whose  vertex  is  at  0. 

The  arcs  A  B,  A  P  and  P  B  are  measures  of  these  A.  §  202 

NowZAOB<A  AOP  +  A  POB,  §487 

(the  sum  of  any  two  face  A  of  a  trihedral  is  >  the  third  Z.). 

.'.  arc  A  B  <  arc  A  P  +  arc  P  B. 

In  like  manner,  joining  any  point  in  A  C  P  with  A  and  P 
by  arcs  of  great  (D,  their  sum  would  be  greater  than  arc  A  P ; 

and,  joining  any  point  in  P  Q  B  with  P  and  B  by  arcs  of 
great  (D,  the  sum  of  these  arcs  would  be  greater  than  arc  P  B. 

If  this  process  be  indefinitely  repeated  the  distance  from  A 
to  B  on  the  arcs  of  the  great  ©  will  continually  increase  and 
approach  to  the  line  A  C  P  Q  B. 

.\a.vcAB<ACPQB. 

Q.  E.  D. 


360 


GEOMETRY. 


BOOK    VIII. 


Proposition  IX.     Theorem. 

708.   Every  point  in  an  arc  of  a  great  circle 
bisects  a  given  arc  at  right  angles  is  equally  distant  from  the 
extremities  of  the  given  arc. 

Let   arc  CD  bisect  arc  A  £  at 
right  angles. 

We  are  to  prove  any  point  0  in 
G  D  is  equally  distant  from  A  and  B. 

Since  great  circle  CDE  bisects 
arc  A  B  at  right  angles,  it  also  bisects 
chord  A  B  at  right  angles. 

Hence,  chord  A  B  is  _L  to  the 
plane  (7  Z>^  at  K. 

.'.  0  K  is  J_  to  chord  A  B  at  its  middle  point. 

.'.straight  lines  0 A  and  OB  are  equal. 


.'.arcs  0  A  and  OB  are  equal. 


§430 

§58 

§182 


Q.  E.  D. 


Proposition  X.     Problem. 

709.  To  pass  the  circumference  of  a  small  circle  through 
any  three  points  on  the  surface  of  a  sphere. 

Let  A,  B  and  C  be  any  three 
points  on  the  surface  of  a 
sphere. 

It  is  required  to  pass  the  circum- 
ference of  a  small  circle  through  the 
points  A,  B  and  C. 

Pass  arcs  of  great  circles  through 
A  and  B,  A  and  C,  B  and  0.       §  704 

Arcs  of  great  circles  a  o  and  b  o 
J_  to  A  C  and  B  C  at  their  middle  points  intersect  at  o. 

Then  o  is  equally  distant  from  A,  B  and  C.  §  708 

.*.  the  circumference  of  a  small  circle  drawn  from  o  as  a 

pole,  with  an  arc  o  A  will  pass  through  A,  B  and  C,  and  be  the 

circumference  required. 

Q.  E.  D. 


THE    SPHERE.  361 


On  Spherical  Angles. 

710.  Def.  The  angle  of  two  curves  which  have  a  common 
point  ia  the  angle  included  by  the  two  tangents  to  the  two  curves 
at  that  point. 

711.  Def.  A  spherical  angle  is  the  angle  included  between 
two  arcs  of  great  circles. 

Proposition  XI.     Theorem. 

712.  The  angle  of  two  curves  which  intersect  on  the  sur- 
face of  a  sphere  is  equal  to  the  dihedral  angle  between  the 
planes  passed  through  the  centre  of  the  sphere,  and  the  tan- 
gents of  the  two  curves  at  their  point  of  intersection. 


Let  the  curves  A  B  and  A  C  intersect  at  A  on  the  sur- 
face of  a  sphere  whose  centre  is  0 ;  and  let  A  T 
and  A  S  be  the  tangents  to  the  two  curves  re- 
spectively. 

We  are  to  prove  Z  T  AS  equal  to  the  dihedral  angle  formed 
by  the  planes  OAT  and  0  A  S. 

Since  A  T  and  A  S  do  not  cut  the  curves  at  A,  they  do  not 
cut  the  surface  of  the  sphere, 

and  are  therefore  tangents  to  the  sphere. 
.'.AT  and  A  S  are  J_  to  the  radius  0  A,  drawn  to  the  point 
of  contact.  §  186 

.*.  Z  T  AS  measures  the  dihedral  Z  of  the  planes  OAT 
and  0  A  S,  passed  through  the  radius  0  A  and  the  tangents  A  T 
and  AS.  §  470 

But  Z  TA  S  is  the  Z  of  the  two  curves  A  B  and  A  C.    §  710 

.'.  the  Z  of  the  two  curves  A  B  and  AC  =  the  dihedral  Z 
of  the  planes  0  A  T  and  0  A  S. 

Q.  E.  D. 


362 


GEOMETRY. 


BOOK    VIII. 


Proposition  XII.     Theorem. 

713.  A  spherical  angle  is  equal  to  the  measure  of  the 
dihedral  angle  included  by  the  great  circles  whose  arcs  form 
the  sides  of  the  angle. 

P 


Let  BPC  be  any  spherical  angle,  and  B P D  P'  and 
C P  E P'  the  great  circles  whose  arcs  B P  and  C P 
include  the  angle. 

We  are  to  prove  /.BPC  equal  to  the  measure  of  the  dihe- 
dral Z  C-PP'-B. 

Since  two  great  ©  intersect  in  a  diameter,  P  P'  is  a 
diameter.  §  685 

Draw  P  T  tangent  to  the  O  BPDP1. 

Then  P  T  lies  in  the  same  plane  as  the  O  B  P  D  P',  and  is 

_L  to  PP<  at  P. 

In  like  manner  draw  P T'  tangent  to  the  O  CPEP'. 
Then  P  T'  lies  in  the  same  plane  as  the  O  C  P  EP',  and  is 

-L  to  PP'  at  P. 

.'.  Z  TPT  is  the  measure  of  the  dihedral  Z  C-PP'-B.  §  470 
But  spherical  Z  B  P  C  is  the  same  as  plane  ZTPT';  §  7 1 0 
.*.  spherical  Z  BPC  is  equal  to  the  measure  of  dihedral 

Z  C-PP'-B. 

Q.  E.  D. 

714.  Corollary.  A  spherical  angle  is  measured  by  the  art 
of  a  great  circle  described  about  its  vertex  as  a  pole  and  intercepted 
by  its  sides  (produced  if  necessary).  For,  if  B  C  be  the  arc  of  a 
great  circle  described  about  the  vertex  P  as  a  pole,  P  B  and  P  C 
are  quadrants.  Hence,  B  0  and  C  0  are  perpendicular  to  P  P'. 
Therefore  BO  C  measures  the  dihedral  angle  B-P  O-C,  and, 
hence,  the  spherical  angle  BPC.  Therefore,  arc  B C,  which 
measures  the  angle  BO C,  measures  the  spherical  angle  BPC. 


THE    SPHERE.  363 


On  Spherical  Polygons  and  Pyramids. 

715.  Def.  A  spherical  Polygon  is  a  portion  of  a  surface  of 
a  sphere  bounded  by  three  or  more  arcs  of  great  circles. 

The  sides  of  a  spherical  polygon  are  the  bounding  arcs ; 
the  angles  are  the  angles  included  by  consecutive  sides;  the 
vertices  are  the  intersections  of  the  sides. 

716.  Def.  The  Diagonal  of  a  spherical  polygon  is  an  arc 
of  a  great  circle  dividing  the  polygon,  and  terminating  in  twG 
vertices  not  adjacent. 

The  planes  of  the  sides  of  a  spherical  polygon  form  by 
their  intersections  a  polyhedral  angle  whose  vertex  is  the  centre 
of  the  sphere,  and  whose  face  angles  are  measured  by  the  sides 
of  the  polygon. 

717.  Def.  A  spherical  Pyramid  is  a  portion  of  a  sphere 
bounded  by  a  spherical  polygon  and  the  planes  of  the  sides  of 
the  polygon. 

The  spherical  polygon  is  the  base  of  the  pyramid,  and  the 
centre  of  the  sphere  is  its  vertex. 

718.  Def.  A  spherical  Triangle  is  a  spherical  polygon  of 
three  sides. 

A  spherical  triangle,  like  a  plane  triangle,  is  right,  or  oblique  ; 
scalene,  isosceles  or  equilateral. 

719.  Def.  Two  spherical  triangles  are  equal  if  their  suc- 
cessive sides  and  angles,  taken  in  the  same  order,  be  equal  each 
to  each. 

720.  Def.  Two  spherical  triangles  are  symmetrical  if  their 
successive  sides  and  angles,  taken  in  reverse  order,  be  equal  each 
to  each. 

721.  Def.  The  Polar  of  a  spherical  triangle  is  a  spherical 
triangle,  the  poles  of  whose  sides  are  respectively  the  vertices  of 
the  given  triangle. 

Since  the  sides  of  a  spherical  triangle  are  arcs,  they  may  be 
expressed  in  degrees  and  minutes. 


364 


GEOMETRY. BOOK   VIII. 


Proposition  XIII.     Theorem. 

722.  Any  side  of  a  spherical  triangle  is  less  than  the 
of  the  other  two  sides. 


Let  ABC  be  any  spherical  triangle. 
We  are  to  prove  BG  <  B  A  +  AG. 
Join  the  vertices  A,  B  and  G  with  the 
centre  0  of  the  sphere. 

Then,  in  the  trihedral  A  O-A  BG  thus 
formed,  the  face  A  A  0  G,  AOB  and 
BOG  are  measured,  respectively,  by  the 
sides  A  G,  A  B  and  B  G.  §  202 

Now,  BOG<BOA  +  AOG,  §  487 

any  two  A  of  a  trihedral  is  greater  than  the  third  Z. ). 

.'.BG<BA  +  AG. 

Q.  E.  D. 


sum  of 


723.  Corollary.   Any  side  of  a  spherical  polygon  is  less 
than  the  sum  of  the  other  sides. 


Ex.  1.  Given  a  cone  of  revolution  whose  side  is  24  feet,  and 
the  diameter  of  its  base  6  feet ;  find  its  entire  surface,  and  its 
volume. 

2.  Given  the  frustum  of  a  cone  whose  altitude  is  24  feet, 
the  circumference  of  its  lower  base  20  feet,  and  that  of  its  upper 
base  16  feet;  find  its  volume. 

3.  The  volume  of  the  frustum  of  a  cone  of  revolution  is 
8025  cubic  inches;  its  altitude  14  inches;  the  circumference  of 
the  lower  base  twice  that  of  the  upper  base.  What  are  the  cir- 
cumferences of  the  bases  1 

4.  The  frustum  of  a  cone  of  revolution  whose  altitude  is 
20  feet,  and  the  diameters  of  its  bases  12  feet  and  8  feet  respec- 
tively, is  divided  into  two  equal  parts  by  a  plane  parallel  to  its 
bases.     What  is  the  altitude  of  each  part  1 


THE    SPHERE. 


365 


Proposition  XIV.     Theorem. 

724.   The  sum  of  the  sides  of  a  spherical  polygon  is  less 
than  the  circumference  of  a  great  circle. 


Let  ABC  BE  be  a  spherical  polygon. 

We  are  to  prove  AB  +  B  G  etc.  less  than  the  circumference 
of  a  great  circle. 

Join  the  vertices  A,  B,  G  etc.,   with  0  the  centre  of  the 
sphere. 

The  sum  of  the  face  A  A  0  B,  BOG  etc.,  which  form  a 
polyhedral  Z  at  0,  is  less  than  four  rt.  A .  §  488 

.*.  the  sum  of  the  arcs  A  B,  B  G  etc.,  which  measure  these 
face  A ,  is  less  than  the  circumference  of  a  great  circle. 

Q.  E.  D. 

725.  Corollary.    If  we  denote  the  sides  of  a  spherical  tri- 
angle by  a,  b  and  c,  then  a  +  b  +  c  <  360°. 


Ex.  1.  Tho  surface  of  a  cone  is  540  square  inches;  what 
is  the  surface  of  a  similar  cone  whose  volume  is  8  times  as 
great  1 

2.  The  lateral  surface  of  a  cone  is  S ;  what  is  the  lateral 
surface  of  a  similar  cone  whose  volume  is  n  times  as  great  ] 


366 


GEOMETRY. BOOK   VIII. 


Proposition  XV.     Theorem. 

726.  A  point  upon  the  surface  of  a  sphere,  which  is  at 
the  distance  of  a  quadrant  from  each  of  two  other  points,  is 
one  of  the  poles  of  the  great  circle  which  passes  through  these 
points. 


Let  P  be  a  point  at  the  distance  of  a  quadrant  from 
each  of  the  two  points  A  and  B. 

We  are  to  prove  P  a  pole  of  the  great  circle  which  passes 
through  A  and  B. 

Since  PA  and  P B  are  quadrants, 

A  POA  and  P  0  B  me  it  A. 

.'.  P  0  is  _L  to  the  plane  of  the  O  ABC,  §  449 

(a  straight  line  _L  to  two  straight  lines  drawn  through  its  foot  in  a  plane  is 
_L  to  the  -plane). 


P  is  a  pole  of  the  O  A  B  G. 


§  683 

Q.  E.  D. 


Ex.  1.  Show  that  two  symmetrical  polyhedrons  may  be  de- 
composed into  the  same  number  of  tetrahedrons  symmetrical  each 
to  each. 

2.  Show  that  two  symmetrical  polyhedrons  are  equivalent. 

3.  Show  that  the  intersection  of  two  planes  of  symmetry  of 
a  solid  is  an  axis  of  symmetry. 

4.  Show  that  the  intersections  of  three  planes  of  symmetry 
of  a  solid  are  three  axes  of  symmetry;  and  that  the  common 
intersection  of  these  axes  is  the  centre  of  symmetry. 


THE    SPHERE. 


367 


Proposition  XVI.     Theorem. 

727.  If,  from  the  vertices  of  a  given  spherical  triangle 
as  poles,  arcs  of  great  circles  be  described,  another  triangle  is 
formed,  the  vertices  of  which  are  the  poles  of  the  sides  of  the 
given  triangle. 


Let  A  B  G  be  the  given  triangle;  and,  from  its  vertices 
A,  B  and  G  as  poles,  let  the  arcs  B'C,  A' C  and 
A' B'  respectively  be  described. 

We  are  to  prove  vertices  A',  B'  and  C  poles  respectively  of 
arcs  BC,  A  G  and  A  B. 

Since  B  is  the  pole  of  the  arc  A'  C',  and  G  the  pole  of  the 
oxcA'B', 

A'  is  at  a  quadrant's  distance  from  each  of  the  points  B  and  G. 

,'J'isa  pole  of  the  arc  B  G,  §  726 

(a  point  upon  the  surface  of  a  sphere  which  is  at  a  quadrant's  distance  from 
each  of  two  other  points  is  one  of  the  poles  of  the  great  circle  which  passes 
through  those  points). 

In  like  manner,  it  may  be  shown  that  B'  is  a  pole  of  the 
arc  A  G,  and  G'  a  pole  of  the  arc  A  B. 

Q.  E.  D. 

728.  Scholium  1.  A  A1  B'  G'  is  the  polar  of  A  A  B  G,  and, 
reciprocally,  A  A  B  C  is  the  polar  of  A  A'  B'  G'. 

729.  Sch.  2.  The  arcs  of  great  circles  described  about  A, 
B  and  C  as  poles  will,  if  produced,  form  three  triangles  exterior 
to  the  polar.  The  polar  triangles  are  distinguished  by  having 
their  homologous  vertices  A  and  A'  on  the  same  side  of  B  G  and 
B1  G',  B  and  B'  on  the  same  side  of  A  G  and  A'  G',  and  G  and 
G'  on  the  same  side  of  A  B  and  A'  B'. 


368  GEOMETRY. BOOK    VIII. 

Proposition  XVII.     Theorem. 
730.  In  two  polar  triangles  each  angle  of  either  is  the 
supplement  of  the  side  lying  opposite  to  it  in  the  other. 


Let  ABG  and  A'  B'  C  be  two  polar  triangles. 

We  are  to  prove  A  A,  B  and  G  respectively  the  supplements 
of  the  sides  B'  C,  A1  C  and  A'  B'. 

Let  the  sides  A  B  and  A  G,  produced  if  necessary,  meet  the 
side  B'  G'  in  the  points  b  and  c. 

Since  the  vertex  A  is  a  pole  of  the  arc  B'  C',        §721 

A  A  is  measured  by  b  c,  §  714 

(a  spherical  Z  is  measured  by  the  arc  of  a  great  circle  described  about  its 
vertex  as  a  pole  and  intercepted  by  its  sides). 

Now,  since  B'  is  the  pole  of  the  arc  Ac,  B'  c  =  90°. 

Since  G'  is  the  pole  of  the  arc  Ab,C'b  =  90°. 

.-.  B'c+G/b  =  B,C,+  bc=l80°. 

.'.  Z  A  ( =  b  c)  is  the  supplement  of  the  side  B'  G'. 

In  like  manner  it  may  be  shown  that  each  A  of  either  A  is 
the  supplement  of  the  side  lying  opposite  to  it  in  the  other. 

Q.  E.  D. 

731.  Scholium.  In  two  polar  triangles  each  side  of  either 
is  the  supplement  of  the  angle  lying  opposite  to  it  in  the  other.  If 
A,  B  and  G  denote  the  angles,  and  a,  b  and  c  the  sides  of  a  tri- 
angle, the  angles  of  the  polar  triangle  will  be  180°  —  a,  180° 
—  b  and  180°  —  c;  and  the  sides  of  the  polar  triangle  will  be 
180°  -  A,  180°  -  B  and  180°  -  G. 

By  reason  of  these  relations  polar  triangles  are  often  called 
supplemental  triangles. 


THE-  SPHERE.  369 


Proposition  XVIII.     Theorem. 

73£.  The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  two,  and  less  than  six,  right  angles. 


Let  ABC  be  a  spherical  triangle. 

We  are  to  prove  ZA+ZB+ZC  greater  than  2,  and  less 
than  6,  right  angles. 

Denote  the  sides  of  the  polar  A  opposite  the  A  A,B,  C  re- 
spectively, by  a',  b',  cf. 

Then  Z  A  =  180°  -a',  Z  B=  180°  -  V  and  Z  C  = 
180°  -  c7,  §  730 

(in  two  polar  A  each  Z  of  either  is  tJie  sitjqrtement  of  the  side  lying  opposite 
to  it  in  the  other.) 

By  adding,     ZA  +  ZB  +  ZC  =  540°  -  (a'  +  V  +  c'). 

But  a'  +  V  +  d  is  less  than  360°,  §  724 

(the  sum  of  the  sides  of  a  spherical  polygon  is  less  than  the  circumference  of 
a  great  circle). 

.'.ZA  +  ZB  +  ZOI&00. 

Also,  since  each  Z  is  less  than  2  rt.  A, 

their  sum  is  less  than  6  rt.  A. 

Q.  E.  D. 

733.  Corollary.  A  spherical  triangle  may  have  two,  or 
even  three  right  angles ;  or  two,  or  even  three  obtuse  angles. 

734.  Def.  A  spherical  triangle  having  one  right  angle  is 
called  rectangular;  having  two  right  angles,  bi-rectangular ; 
having  three  right  angles,  tri-rectangular. 

Each  of  the  sides  of  a  tri-rectangular  triangle  is  a  quadrant, 
and  the  triangle  is  called,  when  reference  is  had  to  its  sides,  tri- 
quadrantal. 


370  GEOMETRY. BOOK    VIII. 

Proposition  XIX.     Theorem. 

735.  Each  angle  of  a  spherical  triangle  is  greater  than 
the  difference  between  two  right  angles  and  the  sum  of  the 
other  two  angles. 


Let  AA,B  and  C  be  the  angles  of  the  spherical  tri- 
angle ABC. 

We  are  to  prove  Z  A  greater  than  the  difference  between  1 80° 
and(ZB  +  ZC). 

I.  Suppose  (Z  B  +  Z  C)  <  180°. 

Now  ZA  +  ZB  +  ZC>  180°.  §  732 

By  transposing,  Z  A  >  180°  -  (Z  B  +  Z  C). 

II.  Suppose  (Z  B  +  Z  C)  >  180°. 

Now  of  the  three  sides  (180°  -  Z  A),  (180°  -  Z  B),  (180° 
—  Z  C),  of  the  polar  A,  each  is  less  than  the  sum  of  the  other 
two,  §  722 

{cither  side  of  a  spherical  A  is  less  than  the  sum  of  the  other  two  sides). 

.'.  (180°  -  Z  B)  +  (180°  -  Z  C)  >  180°  -  Z  A ;     ' 
or,  360°  -  (Z  B  +  Z  C)  >  180°  -  Z  A. 

By  transposing,  Z  A>{ZB  +  ZC)~  180°. 


Q.  E.  D. 


Ex.  1.  The  volume  of  a  cone  is  1728  cuhic  inches;  what  is 
the  volume  of  a  similar  cone  whose  surface  is  4  times  as  great  1 

2.  The  volume  of  a  cone  is  V ;  what  is  the  volume  of  a  simi- 
lar cone  whose  surface  is  n  times  as  great  1 


THE    SPHERE. 


371 


736.  Def.  Equal  spherical  triangles  are  triangles  which 
have  their  corresponding  sides  and  angles  equal  each  to  each  and 
arranged  in  the  same  order,  so  that  when  applied  to  each  other 
they  will  coincide.  Thus  in  Fig.  1,  ABC  and  A'  B'  C  are  equal 
spherical  triangles. 


Fig.  1.  Fig.  2. 

737.  Def.  Symmetrical  spherical  triangles  are  triangles 
which  have  their  corresponding  sides  and  angles  equal  each  to 
each,  but  arranged  in  reverse  order. 

Thus,  in  Fig.  2,  A  B  C  and  A'  B'  C  are  symmetrical  spheri- 
cal triangles.  For,  since  the  face  angles  of  the  two  trihedrals 
are  equal  respectively,  but  are  arranged  in  reverse  order,  the 
sides  of  the  spherical  triangles,  which  measure  these  face  angles, 
are  equal,  each  to  each,  and  are  arranged  in  reverse  order ;  and 
since  the  dihedral  angles  of  the  two  trihedrals  are  equal  respec- 
tively, but  are  arranged  in  reverse  order,  the  angles  of  the 
spherical  triangles,  which  are  equal  to  these  trihedrals,  are  equal, 
each  to  each,  and  are  arranged  in  reverse  order. 

In  like  manner  we  may  have  symmetrical  spherical  poly- 
gons of  any  number  of  sides,  and  corresponding  symmetrical 
spherical  pyramids. 

Two  symmetrical  spherical  triangles  cannot  be  made  to 
coincide.  For,  if  their  convexities  lie  in  opposite  directions, 
they  evidently  will  not  coincide ;  and  if  their  convexities  lie  in 
the  same  direction,  and  we  apply  A  B  to  A'  B',  the  vertices  G 
and  C  will  lie  on  opposite  sides  of  A1  B'. 


372  GEOMETRY. BOOK    VIII. 

738.  There  is,  however,  one  exception.      Two  symmetrical 
isosceles  spherical  triangles  can  be  made  to  coincide. 


Thus,  if  A  B  C  be  an  isosceles  spherical  triangle,  AB  =  AO 
and  in  its  symmetrical  triangle  A1  B'  =  A'  C.  Hence  A  B  = 
A'  C  and  AC  =  A'  B'.  And,  since  A  A  and  A!  are  equal,  if 
A  B  be  placed  on  A'  C,  A  G  will  fall  on  its  equal  A'  B'. 

In  consequence  of  the  relations  established  between  poly- 
hedral angles  and  spherical  polygons,  from  any  property  of  poly- 
hedral angles,  we  may  infer  a  corresponding  property  of  spherical 
polygons.  Reciprocally,  from  any  property  of  spherical  polygons, 
we  may  infer  a  corresponding  property  of  polyhedral  angles. 


Ex.  1.  The  altitude  of  a  cone  of  revolution  is  12  inches ;  at 
what  distances  from  the  vertex  must  three  planes  be  passed  par- 
allel to  the  base  of  the  cone,  in  order  to  divide  the  lateral  surface 
into  four  equal  parts  1 

2.  The  altitude  of  a  given  solid  is  2  inches,  its  surface  24 
square  inches,  and  its  volume  8  cubic  inches ;  find  the  altitude 
and  surface  of  a  similar  solid  whose  volume  is  512  cubic  inches. 

3.  The  volumes  of  two  similar  cones  of  revolution  are  6  cubic 
inches  and  48  cubic  inches  respectively,  and  the  slant  height 
of  the  first  is  5  inches ;  find  the  slant  height  of  the  second. 


THE    SPHERE. 


373 


Proposition  XX.     Theorem. 
739.  Two  symmetrical  spherical  triangles  are  equivalent. 


Let  ABC  and  A1  B'  C  be  two  symmetrical  spherical 
triangles,  having  A  B,  A  C  and  B  C  equal  respectively 
to  A'  B>,  A'C  andB'C. 

We  are  to  prove  A  ABC  ^  A  A'  B'C. 

Let  P  and  P'  be  poles  of  small  circles  which  pass  through 
A,  B,  C  and  A',  B',  C. 

Now,  since  the  arcs  A  B,  A  C  and  B  C  =  A'  B',  A'  C  and 
B'  C  respectively,  the  chords  of  the  arcs  AB,  AC  and  B  C  = 
chords  of  the  arcs  A'  B'}  A'  C  and  B'  C  respectively.  §  181 

.*.  the  plane  A  formed  by  the  chords  of  these  arcs  are 
equal.  §  108 

.*.  ©ABC  and  A1  B' C  which  circumscribe  these  equal 
plane  A  are  equal. 

.*.  the  six  spherical  distances  PA,  P B,  P'  A'  etc.  are  equal, 
{being  polar  distances  of  equal  (D  on  tlie  same  sphere). 

,' .  A  P  A  B,  P'  A'  B'  are  symmetrical  and  isosceles. 

So  likewise  are  A  P  B  C,  P'  B'  C  and  A  PAC,  P'A'  C. 

.'.  A  P  AB  may  be  applied  to  A  P'  A1  B'  and  will  coincide 
with  it.  §  738 

So  likewise  A  PBC  with  A  P'  B'  C  and  A  PAC  with 
A  P'  A'  C. 

.'.  APAB  +  PBC-PAC^AP'A'B'+  P<  B1  C - 
P'A'C. 


.'.A  ABC- A  A' B'C 


Q.  E.  D. 


740.  Corollary.    Two  symmetrical  spherical  pyramids  are 
equivalent. 


374 


GEOMETRY. BOOK    VIII. 


Proposition  XXI.     Theorem. 
741.    On  the  same  sphere,  or  equal  spheres,  two  triangles 
are  either  equal,  or  symmetrical  and  equivalent,  if  two  sides 
and  the  included  angle  of  the  one  be  respectively  equal  to  two 
sides  and  the  included  angle  of  the  other. 


In  the  AABG  and  B E F,  let  Z  A  =  Z  B,  and  the 
sides  A  B  and  A  G  equal  respectively  the  sides 
BE  and  D F. 

We  are  to  prove  A  A  B  C  and  D  E  F  equal,  or  symmetrical 
and  equivalent. 

I.  When  the  parts  of  the  two  A  are  in  the  same  order  as  in  A 

ABC  and  BE F, 

A  A  B  G   can  be   applied   to   A  B  E  F,    as  in  the  corre- 
sponding case  of  plane  A,  and  will  coincide  with  it.      §  106 

II.  When  the  parts  are  in  reverse  order,  as  in  A  A  B  G  and 

B'  E1  F, 

construct  the  A  BE  Asymmetrical  with  respect  to  A  B'E'F. 

Then  A  B  E  F  will  have  its  A  and  sides  equal  respectively 
to  those  of  the  A  B'E'F.  §  737 

Now  in  the  A  A  B  G  and  B  E  F, 

Z  A=Z  B,  AB  =  BE  and  A  G  =  B  F, 

and  these  parts  are  arranged  in  the  same  order. 

.'.  A  A  B  G  =  A  B  EF.  Case  I. 

But  A  B'E'F-  A  BEF,  §  739 

.'.AABG^AB'E'F. 

Q.  E.  D. 


THE    SPHERE. 


m 


Proposition  XXII.     Theorem. 

742.  Two  triangles  on  the  same  sphere,  or  equal  spheres, 
are  either  equal,  or  symmetrical  and  equivalent,  if  a  side  and 
two  adjacent  angles  of  the  one  be  equal  respectively  to  a  side 
and  two  adjacent  angles  of  the  other. 


For  one  of  the  A  may  be  applied  to  the  other,  or  to  its  sym- 
metrical A,  as  in  the  corresponding  case  of  plane  A.  §  107 

Q.  E.  D. 


Proposition  XXIII.     Theorem. 

743.  Two  mutually  equilateral  triangles  on  the  same 
sphere,  or  equal  spheres,  are  mutually  equiangular,  and  are 
either  equal,  or  symmetrical  and  equivalent. 


For  the  face  A  of  the  corresponding  trihedral  angles  at  the 
centre  of  the  sphere  are  equal  respectively,  §  202 

(since  they  arc  measured  by  equal  sides  of  the  A). 
.*.  the  corresponding  dihedral  A  are  equal.  §  492 

.*.  the  A  of  the  spherical  A  are  respectively  equal. 
.'.  the  A  are  either  equal,  or  symmetrical  and  equivalent, 
according  as  their  equal  sides  are  arranged  in  the  same,  or  reverse 
order. 

Q.  E.  D. 


376 


GEOMETRY. BOOK    VIII. 


Proposition  XXIV.     Theorem. 

744.  Two  mutually  equiangular  triangles  on  the  same 
sphere,  or  equal  spheres,  are  mutually  equilateral,  and  are 
either  equal,  or  symmetrical  and  equivalent. 


Let  the  spherical  triangles  ABC  and  D  E  F  be  mutually 
equiangular. 

We  are  to  prove  A  A  B  C  and  DBF  mutually  equilateral, 
and  equal,  or  symmetrical  and  equivalent. 

Let  A  A'  B'  C  and  D'  E'  F  be  the  polar  A  of  A  A  B  0  and 
D  E  F  respectively. 

Then  the  A  A'  B'  C  and  D' E' F  are  mutually  equilat- 
eral, §  731 

(in  two  polar  A  each  side  of  the  one  is  the  supplement  of  the  A  lying  opposite 
to  it  in  the  other). 

.'.  A  A'B'C  and  D E' F  are  mutually  equiangular,  §  743 
(two  mutually  equilateral  A  on  equal  spheres  are  mutually  equiangular). 

.'.  A  A  B  C  and  DEF are  mutually  equilateral ;         §  731 

hence  A  A  B  C  and  D  E  F  are  either  equal,  or  symmetri- 
cal and  equivalent,  §  743 
(two  mutually  equilateral  A  on  equal  spheres  are  either  equal,  or  symmetrical 

and  equivalent). 

Q.  E.  D. 


THE    SPHERE.  377 


Proposition  XXY.     Theorem. 

745.    The  angles   opposite  equal  sides   of  an   isosceles 
spherical  triangle  are  equal. 


In  the  spherical  A  A  B  C,  let  A  B  =  AC. 
We  are  to  prove  Z.  B  =  Z  C. 

Draw  arc  A  D  of  a  great  circle,  from  the  vertex  A  to  the 
middle  of  the  base  B  C. 

Then  A  A  B  D  and  A  C  D  are  mutually  equilateral. 

.'.  A  A  B D  and  A  CD  are  mutually  equiangular,        §  743 
{two  mutually  equilateral  &  on  the  same  sphere  are  mutually  equiangular). 

.-.ZB  =  ZC, 

(since  tliey  are  lwmologous  A  of  symmetrical  &). 

Q.  E.  D. 


746.  Corollary.  The  arc  of  a  great  circle  drawn  from  the 
vertex  of  an  isosceles  spherical  triangle  to  the  middle  of  the  base 
bisects  the  vertical  angle,  is  perpendicular  to  the  base,  and  di- 
vides the  triangle  into  two  symmetrical  triangles. 


378 


GEOMETRY. BOOK  VIII. 


Proposition  XXVI.  ,  Theorem. 

747.    If  two  angles  of  a  spherical  triangle  be  equal,  the 
sides  opposite  these  angles  are   equal,  and   the   triangle  is 


In  the  spherical  A  A  B  C,  let  Z  B  =  Z  C. 


We  are  to  prove  A  C  =  A  B. 

Let       A  A'  B'  a  be  the  polar  AofAi.BC. 

Since  Z  B  =  ZC, 

.\A/C,  =  A'Bf, 
(in  two  polar  A  each  side  of  one  is  the  supplement  of  the  Z 
it  in  the  other). 


Hyp. 
§731 

opposite  to 


.-.  Z  B'  =  Z  C, 


§745 


(in  an  isosceles  spherical  A,  the  A  opposite  the  equal  sides  are  equal). 


AC  =  AB. 


§731 

Q.  E.  D. 


THE    SPHERE.  379 


Proposition  XXVII.     Theorem. 

748.  In  a  spherical  triangle  the  greater  side  is  opposite 
the  greater  angle  ;  and,  conversely,  the  greater  angle  is  oppo- 
site the  greater  side. 


I.  In  the  A  ABC,  let  Z  ABO  Z  0. 
We  are  to  prove  A  C  >  A  B. 

Draw  the  arc  BDofa  great  circle,  making  Z  0  B  D  =  Z  C. 

Then  DC=DB,  §747 

(if  two  A  of  a  spherical  A  be  equal  the  sides  opposite  these  A  are  equal). 

Add  D  A  to  each  of  these  equals ; 

then  DC  +  DA  =DB  +  DA. 

But  DB  +  DA>  AB,  §722 

(the  sum  of  two  sides  of  a  splierical  A  is  greater  tlian  the  third  side). 

.\DC+DA>AB,otAOAB. 

II.  Let  AC  >  A  B. 
We  are  to  prove      Z  A  B  C>  Z  C. 

If  ZABC  =  Z  C,  AC  =  AB,  §747 

andif  Z  ABC<Z  C,AC<AB.  Case  I. 

But  both  of  these  conclusions  are  contrary  to  the  hypothesis. 
.*.  Z  A  B  C>  Z  C. 

Q.  E.  D. 


380  .        GEOMETRY. BOOK    VIII. 

Proposition  XXVIII.     Theorem. 

749.   On  unequal  spheres  mutually  equiangular  triangles 
are  similar. 


From  0,  the  common  centre  of  two  unequal  spheres, 
draw  the  radii  0  A,  0  B  and  0  C  cutting  the  sur- 
face of  the  smaller  sphere  in  a,  b  and  c.      Draw 
arcs  of  great  circles,  AB,  AC,  BC,  ab,  a  c,  be. 
We  are  to  prove      A  AB  C  similar  to  A  ab  c. 

A  A,  B,  C  are  equal  respectively  to  A  a,  b,  c, 
(since  the  corresponding  dihedrals  in  each  case  are  the  same). 

In  the  similar  sectors  A  0  B  and  a  Ob, 

AB  :ab  : :  A  0  :aO;  §385 

and  in  the  similar  sectors  A  0  G  and  aOc, 

AC  :ac::AO  :aO.  §  385 

.*.  A  B  :  ab  ::  A  C  :  ac. 
In  like  manner,  AB  :  ab  : :  B C  :bc. 

That  is,  the  homologous  sides  of  the  two  A  are  proportional, 
and  their  homologous  A  are  equal. 

.'.A  A  B  C  is  similar  to  A  ab  c. 

Q.  E.  D. 

750.  Scholium.  The  statement  that  mutually  equiangular 
spherical  A  are  mutually  equilateral,  and  equal,  or  symmetrical 
and  equivalent,  is  true  only  when  limited  to  the  same  sphere,  or 
equal  spheres.  But  when  the  spheres  are  unequal,  the  spherical 
A  are  similar,  but  not  equal.  Hence,  to  compare  two  similar 
spherical  A,  it  is  necessary  to  know  the  linear  extent  of  two 
homologous  sides  ;  or,  what  is  equivalent,  to  know  the  radii  of 
the  spheres.  And,  as  in  the  case  of  plane  A,  two  similar  spheri- 
cal A  have  the  same  ratio  as  the  squares  of  the  linear  measures 
of  any  two  homologous  sides,  and  therefore  as  the  squares  of  the 
radii  of  the  spheres. 


THE    SPHERE. 


381 


On  Comparison  and  Measurement  of  Spherical  Surfaces. 

751.  Def.    A  Lune  is  a  part  of  the  surface  of  a  sphere  in- 
cluded between  two  semi-circumferences  of  great  circles. 

752.  Def.    The  Angle  of  a  lune  is  ^ 
the  angle  included  by  the  semi-circum- 
ferences   which    forms    its    boundary. 
Thus  Z.  CAB  is  the  angle  of  the  lune. 

753.  Def.  A  Spherical  Ungula,  or 
Wedge,  is  a  part  of  a  sphere  bounded 
by  a  lune  and  two  great  semicircles. 

754.  Def.  The  Base  of  an  ungula 
is  the  bounding  lune. 

755.  Def.  The  Angle  of  an  ungula 
is  the  dihedral  of  its  bounding  semicir- 
cles, and  is  equal  to  the  angle  of  the  bounding  lune. 

756.  Def.    The  Edge  of  an  ungula  is  the  edge  of  its  angle. 

757.  Def.    The  Spherical  Excess  of  a  spherical  triangle  is 
the  excess  of  the  sum  of  its  angles  over  two  right  angles. 

C  758.   Def.    Three  planes  which 

pass  through  the  centre  of  the  sphere, 
each  perpendicular  to  the  other  two, 
divide  the  surface  of  the  sphere  into 
eight  tri-rectangular  triangles.  Thus 
1 5  the  three  planes  0 A  D B,  CEDE 
and  AEBF  divide  the  surface  o£ 
the  sphere  into  the  eight  tri-rectangular 
triangles  C  E  B,  D  E  B,  0  B  E,  DB  F, 
etc. 

As  in  Plane  Geometry  the  whole 
angular  magnitude  about  any  point  in  a  plane  is  divided  by  two 
straight  lines  perpendicular  to  each  other  into  four  right  angles, 
and  each  right  angle  is  measured  by  a  quadrant,  or  fourth  part 
of  a  circumference  described  about  that  point  as  a  centre  with 
any  given  radius ;  so,  if,  through  a  point  in  space,  three  planes 
be  made  to  pass  perpendicular  to  one  another,  they  will  divide 
the  whole  angular  magnitude  about  that  point  into  eight  solid 
right  angles,  each  of  which  is  measured  by  an  eighth  part  of  the 
surface  of  a  sphere  described  about  that  point  with  any  given 
radius. 

And,  as  in  Plane  Geometry,  each  quadrant  which  measures 
a  right  angle  is  divided  into  90  equal  parts  called  degrees,  so 
each  of  the  eight  tri-rectangular  spherical  triangles  is  divided 
into  90  equal  parts  called  degrees  of  surface.  Hence,  the  whole 
surface  of  the  sphere  is  divided  into  720  degrees  of  surface. 


382 


GEOMETRY. 


BOOK    VIII. 


Proposition  XXIX.  Lemma. 
759.  The  area  of  the  surface  generated  by  the  revolution 
of  a  straight  line  about  another  line  in  the  same  plane  with  it 
as  an  axis,  is  equal  to  the  product  of  the  projection  of  the  line 
on  the  axis  by  the  circumference  whose  radius  is  perpendicular 
to  the  revolving  line  erected  at  its  middle  point  arid  termi- 
nated by  the  axis. 


Let  the  straight  line  A  B  revolve  about  the  axis  Y  T 
in  the  same  plane ;  let  E  F  be  its  projection  on 
the  axis;  and  G  0  the  perpendicular  to  A  B  at  its 
middle  point  G,  and  terminated  in  the  axis. 
We  are  to  prove  area  A  B  =  E  F  X  2  it  0  G. 
The  surface  generated  by  A  B  is  the  lateral  surface  of  the 
frustum  of  a  cone  of  revolution. 

Draw  GH±,  and  A  D  II,  to  YY. 

Then  area  A  B  =  A  B  X  2  rr  C  H,  §662 

(the  lateral  area  of  a  frustum  of  a  cone  of  revolution  is  equal  to  the  slant 
height  multiplied  by  the  circumference  of  a  section  equidistant  from  its 
bases). 

The  A  ABD  and  G 0 H are  similar ;  §  287 

.'.AD  :AB  ::  GH  :  GO. 

ButCH  :GO  ::2ttGH  :2ttCO,  §375 

(circumferences  of  ©  have  the  same  ratio  as  their  radii). 

.'.AD  :AB  ::2,rGH'.2irGO. 

.•.ADX2ttGO  =  ABX  2  7rGff. 

.'.  area  ofAB  =  ADX27rCO. 

Now  AD  =  EF.  §  135 

.'.2Lre&AB  =  EFX  2#  GO. 

Q.  E.  D. 

760.  Scholium.  If  either  extremity  of  A  B  he  in  the  axis 
YY',  A  B  generates  the  lateral  surface  of  a  cone  of  revolution ;  and 
if  A  B  be  parallel  to  the  axis  Y  Y',  it  generates  the  lateral  area  of 
a  cylinder  of  revolution.     In  either  case  the  formula  holds  good 


THE    SPHERE. 


Exercises. 

1.  If,  from  the  extremities  of  one  side  of  a  spherical  triangle, 
two  arcs  of  great  circles  be  drawn  to  a  point  within  the  triangle, 
the  sum  of  these  arcs  is  less  than  the  sum  of  the  other  two  sides 
of  the  triangle. 

2.  On  the  same  sphere,  or  on  equal  spheres,  if  two  spherical 
triangles  have  two  sides  of  the  one  equal  respectively  to  two 
sides  of  the  other,  but  the  included  angle  of  the  first  greater 
than  the  included  angle  of  the  second,  then  the  third  side  of  the 
first  will  be  greater  than  the  third  side  of  the  second. 

3.  To  draw  an  arc  perpendicular  to  a  given  spherical  arc, 
from  a  given  point  without  it. 

4.  At  a  given  point  in  a  given  arc,  to  construct  a  spherical 
angle  equal  to  a  given  spherical  angle. 

5.  To  inscribe  a  circle  in  a  given  spherical  triangle. 

6.  Given  a  spherical  triangle  whose  sides  are  60°,  80°,  and 
100°  ;  find  the  angles  of  its  polar  triangle. 

7.  The  volume  of  a  pyramid  is  200  cubic  feet ;  find  the  vol- 
ume of  a  similar  pyramid  which  is  three  times  as  high. 

8.  Find  the  centre  of  a  sphere  whose  surface  shall  pass  through 
three  given  points,  and  shall  touch  a  given  plane. 

9.  Find  the  centre  of  a  sphere  whose  surface  shall  pass  through 
three  given  points,  and  shall  also  touch  the  surface  of  a  given 
sphere. 

10.  Find  the  centre  of  a  sphere  whose  surface  shall  touch  two 
given  planes,  and  also  pass  through  two  given  points  which  lie 
between  the  planes. 


384 


GEOMETRY. BOOK    VIII. 


Proposition  XXX.     Theorem. 

761.    The  area  of  the  surface  of  a  sphere  is  equal  to  the 
product  of  its  diameter  by  the  circumference  of  a  great  circle. 
A  A 


Lei  ABODE  be  the  circumference  of  a  great  circle, 
and  AD  the  diameter,  and  OA  the  radius  of  a 
sphere. 

We  are  to  prove      surface  of  sphere  =  ADX2irOA. 

Let  the  semicircle  and  any  regular  inscribed  semi-polygon 
revolve  together  about  the  diameter  A  D. 

The  semi-circumference  will  generate  the   surface   of  the 
sphere, 

and  the  semi-perimeter  a  surface  equal  to  the  sum  of  the 
surfaces  generated  by  the  sides  A  B,  B  0,  CD,  etc. 

Draw  from  the  centre  0,  _k  0  ff,  0 1  and  0  K  to  the  chords 
AB,BG,  CD,  etc. 


These  J?  bisect  the  chords  and  are  equal ; 
.'.  area  AB  =  AP  X  2  ir  Off; 

area  BC  =  PR  X  2  tt  01; 
and  area  CD=RDX2ttOK. 


§185 
§  759 


THE    SPHERE.  385 


Adding,  and  observing  that  0  H,  0 1  and  0  K  are  equal, 

area,  ABCD  =  (A  P+PR  +  RD)X2nOH. 
.'.area  ABC  D  =  AD  X  2  tt  OH. 

Now,  if  the  number  of  sides  of  the  regular  inscribed  semi- 
polygon  be  indefinitely  increased,  the  surface  generated  by  tho 
semi-perimeter  will  approach  the  surface  of  the  sphere  as  its 
limit,  and  0  H  will  approach  0  A  as  its  limit. 

.'.at  the  limit  we  have 

surface  of  the  sphere  =  ADX2nOA.  §199 

Q.  E.  D. 


762.  Corollary  1.  If  7?  denote  the  radius  of  the  sphere, 
then  A  D  will  equal  2  R,  and  0  A  will  equal  R.  Hence  the 
surface  of  a  sphere  equals  2  R  X  2  it  R  =  4  ir  R2. 

763.  Cor.  2.  Since  the  area  of  a  great  circle  of  a  sphere  is 
equal  to  n  R2  (§  381),  and  the  area  of  the  surface  of  a  sphere  is 
equal  to  4  n  R2,  the  surface  of  a  sphere  is  equal  to  four  great 
circles. 

764.  Cor.  3.  If  we  denote  the  surfaces  of  two  spheres  by 
S  and  aS7,  and  their  radii  by  R  and  R'}  we  have  £  :  S'  : :  4  it  R2  : 
4  ir  R'2,  or  S  :  S'  : :  R2  :  R'2 ;  that  is,  the  surfaces  of  two  spheres 
have  the  same  ratio  as  the  squares  on  their  radii. 

765.  Cor.  4.  Since  S  =  4  n  R2  =  *r  (2  R)\  the  surface  of  a 
sphere  is  equivalent  to  a  circle  whose  radius  is  equal  to  the  diameter 
of  the  sphere. 


386 


GEOMETRY. 


BOOK    VIII. 


Proposition  XXXI.     Theorem. 

766.    A  lune  is  to  the  surface  of  the  sphere  as  the  angle 
of  the  lune  is  to  four  right  angles. 


Let  L  denote  the  lune  AB  EG  whose  angle  is  A;  S, 
the  surface  of  the  sphere;  and  B G D F,  a  great 
circle  whose  pole  is  A. 

L         A 

S~ 4  rt.  A' 


We  are  to  x>rove 


Now  the  arc  B  G  measures  the  Z  A  of  the  lune  ;  §  714 

and  the  circumference  B  G  D  F  measures  4  rt.  A . 

Case  I.  —  If  BO  and  B  CDF  be  commensurable. 

Find  a  common  measure  of  B  C  and  BC D F. 

Suppose  this  common  measure  to  be  contained  in  BG  3  times, 

and  m  BGDF  25  times. 


Then 


4  rt.  A 


/    BG    \  _  3 

\BGD  F/~  25' 


Pass  arcs  of  great  ©  through  A  and  these  points  of  division. 
The  entire  surface  will  be  divided  into  25  equal  lunes,  of 
which  lune  L  will  contain  3. 

"  S  ~  25  * 
A  ^  .  L  A 

4  rt.  A  ~  25  '  "  S  ~  4  rt.  A ' 


But 


THE    SPHERE.  387 


Case  II.  —  If  BO  and  B  CD  F  be  incommensurable, 

the  proposition  can  be  proved  by  the  method  of  limits,  as 
employed  in  §  201. 

Q.  E.  D. 


767.  Corollary.  If  we  denote  the  surface  of  the  tri-rectan- 
gular  triangle  by  T,  the  surface  of  the  whole  sphere  will  be  8  T 
(§  758),  aud  if  we  denote  the  surface  of  the  lune  by  L,  and  its 
angle  by  A,  the  unit  of  the  angle  being  a  right  angle,  we  shall 

have  -—-„  =  -•     Therefore  L  =  T  X  2  A. 

And  if  we  take  the  tri-rectangular  triangle  as  the  unit  of 
surface  in  comparing  surfaces  on  the  same  sphere,  we  shall  have 
L  =  2  A.  That  is,  if  a  right  angle  be  the  unit  of  angles  and  the 
tri-rectangular  triangle  be  the  unit  of  spherical  surfaces,  the  area 
of  a  lune  is  expressed  by  twice  its  angle. 

768.  Scholium.  We  mag  also  obtain  the  area  of  a  lune 
whose  angle  is  known,  on  a  given  sphere,  by  finding  the  area  of  the 
sphere,  and  multiplying  this  area  by  the  ratio  of  the  angle  of  the 
lune,  expressed  in  degrees,  to  360°.  Thus,  if  the  angle  of  the  lune 
be  60°,  the  area  of  the  lune  will  be  ^^  of  the  area  of  the  sphere. 


Ex.  1.  Given  the  radius  of  a  sphere  is  10  feet;  find  the  area 
of  a  lune  whose  angle  is  30°. 

2.  Given  the  diameter  of  a  sphere  is  1 6  feet ;  find  the  area 
of  a  lune  whose  angle  is  75°. 

3.  Given  the  diameter  of  a  sphere  is  20  inches;  find  the 
entire  surface  of  its  circumscribed  cylinder ;  and  of  its  circum- 
scribed cone,  the  vertical  angle  of  the  cone  being  60°. 


388  GEOMETRY. BOOK   VIII. 


Proposition  XXXII.     Theorem. 

769.  If  two  circumferences  of  great  circles  intersect  on 
the  surface  of  a  hemisphere,  the  su?n  of  the  opposite  triangles 
thus  formed  is  equivalent  to  a  lune  whose  angle  is  equal  to 
that  included  by  the  semi-circumferences. 


Let  the  semi- circumferences  BAD  and  GAE  intersect 
at  A   on  the  surface  of  a  hemisphere. 

We  are  to  p?we  A  A  B  G  +  A  DAE  equivalent  to  a  lune 
whose  angle  is  BAG. 

The  semi-circumferences  produced  intersect  on  the  opposite 
hemisphere  at  A'. 

Then  each  of  the  arcs  A  D  and  A'  B  is  the  supplement  of 
AB, 

(hvo  great  ©  bisect  each  other). 

.-.  AD  =  A' B. 

In  like  manner,  A  E  —  A1 '0  and  D  M  -■  B  G. 

.'.  A  A  D E  and  A' BG  are  symmetrical  and  equiva- 
lent. §  743 

.\AABG+AADE  =  AABG+  A  A'  B  C  =  lune 
ABA'GA. 

That  is,  A  ABC  +  A  ADE  =  lune  whose  Z  is  BA  G. 

Q.  E.  D. 

770.  Corollary.  The  sum  of  two  spherical  pyramids,  the 
sum  of  whose  bases  is  equivalent  to  a  lune,  is  equivalent  to  a 
wedge  whose  base  is  the  lune. 


THE    SPHERE. 


Proposition  XXXIII.     Theorem. 
771.    The  area  of  a  spherical  triangle  is  equal  to  the 
tri-rectangular  triangle  multiplied  ly  the  ratio  of  the  spherical 
excess  of  the  given  triangle  to  one  right  angle. 


Let  ABC  be  a  spherical  triangle,  and  T  the  area  of 
the  tri-rectangular  triangle. 
We  are  to  prove     AABC  =  T(AA  +  B+C  —  2). 

Complete  the  circumference  A  B  D  E. 
Produce  A  C  and  B  C  to  meet  this  circumference  in  D  and  E. 
Then  A  ABC  +  BCD  (=  lune  A)  =  TX  2  Z  A.    §  767 
AABC  +  ACE(=  lune  B)  =  T  X  2  Z  B,  §  767 

A  A  B  C  +  I)  C  E  (=luue  C)  (§  769)  =  TX  2  Z  C.    §767 
By  adding  these  equalities, 
2AABC+AABC+BCD+ACE+DCE 
=  TX2(AA  +  B+C). 
But     AABC  +  BCIJ  +  ACE+  D  C  E  =  \  T,    §758 

(the  surface  of  a  hemisphere  is  equal  to  4  tri-rectcouju,lar  &). 

.-.  2  A  ABC  +  4  T=  TX2  (AA+  B+  C); 

.- .  A  A  B C  =  T  X  (A  A  +  B  +  C  -  2). 

Q.  E.  D. 

772.  Scholium  1.  If  Z  A  =  140°,  Z^=  120°  and  Z  C  = 
1 00°,  a  right  angle  being  the  unit, 

then,  A^C=  W!ii° +i^!  +  1^!  -  2  W  2  ?'. 
\90°        90°         90°  / 

773.  Scho.  2.  To  find  the  area  of  a  spherical  triangle  on  a 
given  sphere,  the  angles  of  the  triangle  being  given,  we  may  multi- 
ply the  area  of  the  hemisp/iere  by  the  ratio  of  the  spheHcal  excess 
to  360°. 

Thus  if  Z  A  =  140°,  Z  B  =  120°  and  Z  C  =  100°,  since 
the  hemisphere  is  2  n  R\  we  have  AABC  =  2  n  R2  X 
Z  A  +  Z  B  +  Z  C-  180°  )2      18<T  _ 

360°   "  "  2  ""  7l>2  X  3605  ~"  v  ^ 


390 


GEOMETRY. BOOK   VIII. 


Proposition  XXXIV.     Theorem. 
774.    The  area   of  a  spherical  polygon  is  equal  to  ike 
tri-rectangular  triangle  multiplied  by  the  ratio  of  the  spherical 
excess  to  one  right  angle. 


Let  P  denote  the  area,  of  the  spherical  polygon ;  S  the 
sum  of  its  angles;  n  the  number  of  its  sides ;  t,  t', 
t"  .  .  .  the  areas  of  the  triangles  formed  by  drawing 
diagonals  from  any  vertex  A  ;  s,  s'}  s"  ...  respec- 
tively the  sums  of  the  angles  of  these  triangles; 
and  T  the  tri- rectangular  triangle. 

We  are  to  prove         P  =  T  [S  —  2  (n  —  2)  ]. 

Now  t  =  T  (s  -  2),  §  771 

(the  area  of  a  spherical  A  is  equal  to  its  spherical  excess  multiplied  into  the 
area  of  the  tri-rectangular  A). 

t'  =  T  (J  -  2),  §  771 

and  t"  =  T  (s"  —  2),  .  .  . 

By  adding  these  equalities, 

t  +  t'  +  t",  .  .  .  —  T  [s  +  s*  +  s"  +  .  .  .  -  2  (n  -  2)  ]. 

But  t  +  t'  +  t"  +  .  .  .  =  P-, 

and  s  +  s'  +  s"  +  .  .  .  =  8. 

,'.f  =  T[S-2  {n-2)]. 


THE    SPHERE. 


39J 


775.  Corollary.  The  volume  of  a  spherical  pyramid  is  to 
the  volume  of  the  tri-rectangular  pyramid,  as  the  base  of  the  pyra- 
mid is  to  the  tri-rectangular  triangle.  And,  since  the  volume  of 
the  tri-rectangular  pyramid  is  |-  the  volume  of  the  sphere,  and 
the  area  of  the  tri-rectangidar  triangle  is  ^  of  the  surface  of  the 
sphere ;  the  volume  of  a  spherical  pyramid  is  to  the  volume  of  the 
sphere  as  its  base  is  to  the  surface  of  the  sphei*e. 


776.  Def.  A  Zone  is  the  part  of  the  surface  of  a  sphere  in- 
cluded between  two  parallel  circles  of  the  sphere ;  as  the  surface 
included  between  the  circles  ABC  and  E FG. 

777.  Def.  The  Bases  of  a  zone  are  the  circumferences  of 
the  intercepting  circles;  as  circumferences  ABC  and  EFG. 
If  the  plane  of  one  base  become  tangent  to  the  sphere,  that 
base  becomes  a  point,  and  the  zone  will  have  but  one  base. 

778.  Def.  The  altitude  of  a  zone  is  the  perpendicular  dis- 
tance between  the  planes  of  its  bases. 

779.  Def.  A  Spherical  Segment  is  a  part  of  the  sphere  in- 
cluded between  two  parallel  planes. 

780.  Def.  The  Bases  of  a  spherical  segment  are  the  bound- 
ing circles. 

One  of  the  planes  may  become  a  tangent  plane  to  the  sphere. 
In  this  case  the  segment  has  but  one  base. 

781.  Def.  The  Altitude  of  a  spherical  segment  is  the  per- 
pendicular distance  between  the  planes  of  its  bases. 


392 


GEOMETRY. BOOK   VIII. 


782.  Def.  A  Spherical  Sector  is  a  part  of  a  sphere  gener- 
ated by  a  circular  sector  of  the  semicircle  which  generates  the 
sphere  ;  as  A  0  C  K. 

783.  Def.  The  Base  of  a  spherical  sector  is  the  zone  gener- 
ated by  the  arc  of  the  circular  sector ;  as  AC  K. 

The  other  bounding  surfaces  of  a  spherical  sector  may  be 
one  conical  surface,  or  two  conical  surfaces ;  or  one  conical  and 
one  plane  surface. 

Thus,  let  A  B  be  the  diameter  around  which  the  semicircle 
AG  B  revolves  to  generate  the  sphere.  The  solid  generated  by 
the  circular  sector  A  0  G  will  be  a  spherical  sector  having  the 
zone  AG  K  for  its  base,  and  for  its  other  bounding  surface  the 
conical  surface  generated  by  CO. 

The  spherical  sector  generated  by  C  0  D  has  for  its  base  the 
zone  generated  by  G  D,  and  for  its  other  surfaces  the  concave 
conical  surface  generated  by  D  0,  and  the  convex  conical  surface 
generated  by  G  0. 

The  spherical  sector  generated  by  E  0  F  has  for  its  base  the 
zone  generated  by  E  Ft  and  for  one  surface  the  plane  surface 
generated  by  E  0,  and  for  the  other  surface  the  concave  conical 
surface  generated  by  FO. 


THE    SPHEKE. 


393 


Proposition  XXXV.     Theorem. 

784.    The  area  of  a  zone  is  equal  to  the  product  of  its 
altitude  by  the  circumference  of  a  great  circle. 


Let  A  BC D  E  be  the  circumference  of  a  great  circle, 
BC  any  arc  of  this  circumference,  and  0  A  the 
radius  of  the  sphere.  And,  let  PR  be  the  altitude 
of  the  zone  generated  by  arc  B  C. 

We  are  to  prove    zone  B  C  =  P  R  X  2  tt  0  A. 

If  the  semicircle  A  BC D  revolve  about  the  diameter  A  D 
as  an  axis,  the  semi-circumference  ABC D  will  generate  the  sur- 
face of  a  sphere  ;  the  arc  B  C,  a  zone, 

and  the  chord  B  (7,  a  surface  whose  area  is  PR  X  2  it  0 1.  §  759 

Now  if  we  bisect  the  arc  B  C,  and  continue  this  process  in- 
definitely, the  surface  generated  by  the  chords  of  these  arcs  will 
approach  the  zone  as  its  limit ; 

the  _L  0 1  will  approach  the  radius  of  the  sphere  as  its  limit ; 

while  P  R  will  remain  constant. 

.-.  at  the  limit,  zone  BC  =  PRX2nOA. 

Q.  E.  D. 

785.  Corollary  1.  Zones  on  the  same  sphere,  or  equal 
spheres,  have  the  same  ratio  as  their  altitudes. 

786.  Cor.  2.  A  zone  is  to  the  surface  of  the  sphere  as  the 
altitude  of  the  zone  is  to  the  diameter  of  the  sphere. 

787.  Cor.  3.  Let  arc  A  B  generate  a  zone  of  a  single  base. 
Then,  zone  AB_  =  A  P  X  2  tt  0  A.  Hence,  zone  AB  =  ir  AP 
X  AD  =  tt  A~B2.  (§  307.)  That  is,  a  zone  of  one  base  is  equiv- 
alent to  a  circle  whose  radius  is  the  chord  of  the  generating  arc. 


394 


GEOMETRY. 


BOOK    VIII. 


On  the  Volume  of  the  Sphere. 


Proposition  XXXVI.     Theorem. 

788.   The  volume  of  a  sphere  is  equal  to  the  area  of  its 
■nrface  multiplied  by  one-third  of  its  radius. 


Let  R  be  the  radius  of  a  sphere  whose  centre  is  0,  S  its 
surface,  and  V  its  volume. 

We  are  to  prove  V  =  S  X  ^  R. 

Conceive  a  cube  to  be  circumscribed  about  the  sphere. 

From  0,  the  centre  of  the  sphere,  conceive  lines  to  be 
drawn  to  the  vertices  of  each  of  the  polyhedral  AA,B,C,D,  etc. 

These  lines  are  the  edges  of  six  quadrangular  pyramids, 
whose  bases  are  the  faces  of  the  cube,  and  whose  common  altitude 
is  the  radius  of  the  sphere. 

The  volume  of  each  pyramid  is  equal  to  the  product  of  its 
base  by  £  its  altitude.  §  574 

.*.  the  volume  of  the  six  pyramids,  that  is,  the  volume  of 
the  circumscribed  cube,  is  equal  to  the  surface  of  the  cube  mul- 
tiplied by  \  R. 

Now  conceive  planes  drawn  tangent  to  the  sphere,  cutting 
each  of  the  polyhedral  A  of  the  cube. 

We  shall  then  have  a  circumscribed  solid  whose  volume  will 
be  nearer  that  of  the  sphere  than  is  the  volume  of  the  circum- 
scribed cube. 


THE    SPHERE.  395 


From  0  conceive  lines  to  be  drawn  to  each  of  the  polyhedral 
A  of  the  solid  thus  formed,  a,  b,  c,  etc. 

These  lines  will  form  the  edges  of  a  series  of  pyramids, 
whose  bases  are  the  surface  of  the  solid,  and  whose  common  alti- 
tude is  the  radius  of  the  sphere ; 

and  the  volume  of  each  pyramid  thus  formed  is  equal  to 
the  product  of  its  base  by  J  its  altitude. 

.'.  the  sum  of  the  volumes  of  these  pyramids,  that  is,  the 
volume  of  this  new  solid,  is  equal  to  the  surface  of  the  solid  mul- 
tiplied by  J  R. 

Now,  this  process  of  cutting  the  polyhedral  A  by  tangent 
planes  may  be  considered  as  continued  indefinitely, 

and,  however  far  this  process  is  carried,  it  will  always  be 
true  that  the  volume  of  the  solid  is  equal  to  its  surface  multiplied 

But  the  sphere  is  the  limit  of  this  circumscribed  solid. 

.\  V=SX}R.  §  199 

Q.  E.  D. 

789.  Corollary  1.  Since  £=4  *  R2  (j  7C2),  F=4Wi?2X 
l7?=  +  7r/r8.      If  we   denote   the   diameter  of  the  sphere  by 

*.*-(T)-T"rTi»* 

790.  Cor.  2.  Denote  the  radius  of  another  sphere  by  R'  and 
its  volume  by  V  :  we  have  V'=  4  ir  R/S.     .'•  -p-,  =  ^ — tt^==  t^t- 

J     '  J  I'     $nR/a    R'8 

That  is,  spheres  are  to  each  other  as  the  cubes  of  their  radii. 

791.  Cor.  3.  The  volume  of  a  spherical  sector  is  equal  to  the 
product  of  the  area  of  the  zone  which  forms  its  base  by  one-third 
the  radius  of  the  sphere. 

Let  R  denote  the  radius  of  a  sphere,  C  the  circumference  of 
a  great  circle,  H  the  altitude  of  the  zone,  Z  the  surface  of  the 
zone,  and  V  the  volume  of  the  corresponding  sector. 


396 

GEOMETRY. BOOK    VIII. 

Then 

<7  =  2tt7?; 

§  381 

Z=0  X  11=2  w  RX  H; 

§  784 

V=lZXR  =  l-nR2XH. 

792.  Cor.  4.  The  volumes  of  spherical  sectors  of  the  same 
sphere,  or  equal  spheres,  are  to  each  other  as  the  zones  which  form 
their  bases,  or  as  the  altitudes  of  these  zones. 

For,  let  V  and  V  denote  the  volumes  of  two  spherical 
sectors,  Z  and  Z'  the  zones  which  form  their  bases,  H  and  W 
the  altitudes  of  these  zones,  and  R  the  radius  of  the  sphere. 


Then 


And  since 


V  = 

=  Z  X 
Z'  X 

\R_ 

Z' 

z  _ 

H 

Z' 

w 

V  _ 

_H 

V 

H' 

§785 


793.  Cor.  5.  The  volume  of  a  spherical  segment  of  one 
base,  less  than  a  hemisphere,  generated  by  the  revolution  of  a 
semi-segment  ABC  about  the  diameter  A  D,  may  be  found  by 
subtracting  the  volume  of  the  cone  of  revolution  generated  by 
0  B  G  from  that  of  the  spherical  sector  A  0  B. 

In  like  manner,  the  volume  of  a  spherical  segment  of  one 
base,  greater  than  a  hemisphere,  generated  by  the  revolution  of 


THE    SPHERE.  897 


A  B'C  may  be  found  by  adding  the  volume  of  the  cone  of  revo- 
lution generated  by  0  B'  C  to  that  of  the  spherical  sector  gener- 
ated by  A  0  B'. 

794.  Cor.  6.  The  volume  of  a  spherical  segment  of  two 
bases,  generated  by  the  revolution  of  C  B  B'  C  about  the  diame- 
ter A  D,  may  be  found  by  subtracting  the  volume  of  the  segment 
of  one  base  generated  by  A  B  C  from  that  of  the  segment  of 
one  base  generated  by  A  B'  C 


Exercises. 


1.  Given  a  sphere  whose  diameter  is  20  inches;  find  the  cir- 
cumference of  a  small  circle  whose  plane  cuts  the  diameter  4 
inches  from  the  centre. 

2.  Construct,  on  the  spherical  blackboard,  spherical  angles  of 
30°,  45°,  90°,  120°,  150°  and  135°. 

3.  Construct,  on  the  spherical  blackboard,  a  spherical  triangle, 
whose  sides  are  100°,  80°  and  70°  respectively.  What  is  true 
of  its  polar  triangle  1 

4.  Find  the  surface  and  volume  of  a  sphere  whose  radius  is  10 
inches ;  also  find  the  area  of  a  spherical  triangle  on  this  sphere, 
the  angles  of  the  triangle  being  80°,  85°  and  100°  respectively. 

5.  If  7  equidistant  planes  cut  a  sphere,  each  perpendicular  to 
the  same  diameter,  what  are  the  relative  areas  of  the  zones? 

6.  Given,  two  mutually  equiangular  triangles  on  spheres  whose 
radii  are  10  inches  and  40  inches  respectively ;  what  are  their 
relative  areas  ? 

7.  Let  V  denote  the  volume  of  a  spherical  pyramid,  S  its  base, 
E  the  spherical  excess  of  its  base,  and  R  the  radius  of  the  sphere ; 
show  that  S  =  £  7T  R2  E,  and  V  =  £  *r  R*  E. 

8.  Given,  the  volume  of  a  sphere  1728  inches  :  find  its  radius 


GEOMETRY. BOOK    VIII. 


9.  Find  the  ratio  of  the  surfaces,  and  the  ratio  of  the  volumes, 
of  a  cube  and  of  the  inscribed  sphere. 

10.  Find  the  ratio  of  the  surfaces,  and  the  ratio  of  the  vol- 
umes, of  a  sphere  and  the  circumscribed  cylinder. 

11.  Let  V  denote  the  volume  and  //  the  altitude  of  the  spher- 
ical segment  of  one  base,  and  R  the  radius  of  the  sphere ;  show 
tb»t   V=n  IP  (R  -  (  II).      Also,  find    V  when  R  =  12  and 

12.  Given,  a  sphere  2  feet  in  diameter;  find  the  volume  of  a 
segment  of  the  sphere  included  between  two  parallel  planes,  one 
at  3  and  the  other  at  9  inches  from  the  centre.     (Two  solutions.) 

13.  A  sphere  4  inches  in  diameter  is  bored  through  the  centre 
with  a  two-inch  auger  j  find  the  volume  remaining. 


THE    END. 


Presswork  by  Berwick  A  Smith,  J 18  Purchase  Street,  Boston.. 


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is  by  solving  problems ;  not  by  memorizing  rules,  or  solving  prop- 
ositions. Hence  stereotype  methods  and  set  rules  are  avoided. 
Such  problems  are  selected  as  are  calculated  to  interest  the  pupil 
and  lead  him  to  independent  thought  and  discovery.  The  prob- 
lems cover  a  wide  range  of  subjects,  and  are  particularly  adapted 
to  general  mental  discipline,  to  preparation  for  higher  studies, 
mechanical  work,  business  or  professional  life. 


2 

Decimal  fractions  are  introduced  at  an  early  stage,  and  abundant 
practice  in  operations  with  them  is  given  by  means  of  the  metric 
system.  The  chapter  on  the  metric  system  may  be  omitted  without 
affecting  the  unity  of  the  book  ;  but  teachers,  even  if  opposed  to  the 
substitution  of  the  metric  for  the  ordinary  measures,  can  use  this 
chapter  to  great  advantage  as  a  drill  in  the  decimal  system.  Ex- 
perience has  shown  that  the  best  preparation  for  learning  common 
fractions  and  the  common  measures  is  a  thorough  familiarity  with 
decimals. 

Percentage,  in  its  various  applications,  is  fully  explained,  and  is 
illustrated  by  many  examples,  so  that  the  pupil  will  understand  the 
principles  involved,  and  work  intelligently  in  after  life,  whether  he 
is  required  to  compute  interest,  average  accounts,  etc.,  directly,  or 
by  means  of  interest  tables. 

The  nature  and  use  of  logarithms  are  briefly  treated,  and  a  four- 
place  table  of  the  logarithms  of  the  natural  numbers  from  1  to  1,000 
is  given  for  the  purpose  of  saving  time  and  labor  in  the  solution  of 
many  practical  questions.  The  general  method  of  approximations 
is  explained  and  made  very  simple  by  the  use  of  logarithms. 


Wentworth's  Elementary  Algebra. 

Introduction,  $1.12;  Allowance  for  old  book  in  use,  40  cts. 

This  book  is  designed  for  high  schools  and  academies,  and  con- 
tains an  ample  amount  for  admission  to  any  college. 

The  single  aim  in  writing  this  volume  has  been  to  make  an 
Algebra  which  the  beginner  would  read  with  increasing  interest, 
intelligence,  and  power.  The  fact  has  been  kept  constantly  in  mind 
that,  to  accomplish  this  object,  the  several  parts  must  be  presented 
so  distinctly  that  the  pupil  will  be  led  to  feel  that  he  is  mastering 
the  subject.     Originality  in  a  text-book  of  this  kind  is  not  to  be 


expected  or  desired,  and  any  claim  to  usefulness  must  be  based 
upon  the  method  of  treatment  and  upon  the  number  and  character 
of  the  examples.  About  four  thousand  examples  have  been  se- 
lected, arranged,  and  tested  in  the  recitation-room,  and  any  found 
too  difficult  have  been  excluded  from  the  book.  The  idea  has  been 
to  furnish  a  great  number  of  examples  for  practice,  but  to  exclude 
complicated  problems  that  consume  time  and  energy  to  little  or  no 
purpose. 

In  expressing  the  definitions,  particular  regard  has  been  paid  to 
brevity  and  perspicuity.  The  rules  have  been  deduced  from  pro- 
cesses immediately  preceding,  and  have  been  written,  not  to  bo 
committed  to  memory,  but  to  furnish  aids  to  the  student  in  framing 
for  himself  intelligent  statements  of  his  methods.  Each  principle 
has  been  fully  illustrated,  and  a  sufficient  number  of  problems  has 
been  given  to  fix  it  firmly  in  the  pupil's  mind  before  he  proceeds  to 
another.  Many  examples  have  been  worked  out  in  order  to  exhibit 
the  best  methods  of  dealing  with  different  classes  of  problems  and 
the  best  arrangement  of  the  work ;  and  such  aid  has  been  given  in 
the  statement  of  problems  as  experience  has  shown  to  be  necessary 
for  the  attainment  of  the  best  results.  General  demonstrations 
have  been  avoided  whenever  a  particular  illustration  would  serve 
the  purpose,  and  the  application  of  the  principle  to  similar  cases 
was  obvious.  The  reason  for  this  course  is,  that  the  pupil  must 
become  familiar  with  the  separate  steps  from  particular  examples, 
before  he  is  able  to  follow  them  in  a  general  demonstration,  and  to 
understand  their  logical  connection. 

Wentworth's  Complete  Algebra. 

Introduction,  $1.40 ;  Allowance  for  old  book  in  use,  40  cts. 
This  work    is   the    continuation    of  the   Elementary   Algebra 
(described  above),  and  contains  about  150  pages  more  than  that. 


The  additions  are  chapters  on  Chance,  Interest  Formulas,  Contin- 
ued Fractious,  Theory  of  Limits,  Indeterminate  Coefficients,  the 
Exponential  Theorem,  the  Differential  Method,  the  Theory  of 
Numbers,  Imaginary  Numbers,  Loci  of  Equations,  Equations  in 
General,  and  Higher  Numerical  Equations. 


Wentworth  &  McLellan's  Uniuersity  Algebra, 


Wentworth*  s  Plane  Geometry. 

Introduction,  75cts.;  Allowance  for  old  book  in  use,  25cts. 

Wentworth 's  Plane  and  Solid  Geometry. 

Introduction,  $1.25 ;  Allowance  for  old  book  in  use,  40  cts. 

This  work  is  based  upon  the  assumption  that  Geometry  is  a 
branch  of  practical  logic,  the  object  of  which  is  to  detect,  and  state 
precisely  the  successive  steps  from  premise  to  conclusion. 

In  each  proposition,  a  concise  statement  of  what  is  given  is 
printed  in  one  kind  of  type,  of  what  is  required  in  another,  and  the 
demonstration  in  still  another.  The  reason  for  each  step  is  indi- 
cated in  small  type,  between  that  step  and  the  one  following,  thus 
preventing  the  necessity  of  interrupting  the  process  of  demonstra- 
tion by  referring  to  a  previous  proposition.  The  number  of  the 
section,  however,  on  which  the  reason  depends,  is  placed  at  the 
side  of  the  page ;  and  the  pupil  should  be  prepared,  when  called 
upon,  to  give  the  proof  of  each  reason. 

A  limited  use  has  been  made  of  symbols,  wherein  symbols  stand 
for  words,  and  not  for  operations. 


Great  pains  have  been  taken  to  make  the  page  attractive.  The 
propositions  have  been  so  arranged  that  in  no  case  is  it  necessary 
to  turn  the  page  in  reading  a  demonstration. 

A  large  experience  in  the  class-room  convinces  the  author  that, 
if  the  teacher  will  rigidly  insist  upon  the  logical  form  adopted  in 
this  work,  the  pupil  will  avoid  the  discouraging  difficulties  which 
usually  beset  the  beginner  in  geometry;  that  he  will  rapidly  develop 
his  reasoning  faculty,  acquire  facility  in  simple  and  accurate  expres- 
sion, and  lay  a  foundation  of  geometrical  knowledge  which  will  be 
the  more  solid  and  enduring  from  the  fact  that  it  will  not  rest  upon 
an  effort  of  the  memory  simply. 

Went  worth's   Plane   and   Solid  Geometry,    and 

Plane  Trigonometry. 
Introduction,  $1.40;  Allowance  for  old  book  in  use,  40  cts. 


Wentworth's  Plane  Trigonometry. 

Paper.     Introduction,  30  cts. 

Wentworth's  Plane  Trigonometry  and  Logarithms. 

Introduction,  60  cts. 

Wentworth's  Plane  and  Spherical  Trigonometry. 

Introduction,  75  cts.;  Allowance  for  old  book  in  use,  25  cts. 

As  this  work  is  intended  for  beginners,  an  effort  has  been  made 
to  develop  the  subject  in  the  most  simple  and  natural  way. 

In  the  first  chapter,  the  functions  of  an  acute  angle  are  defined 
as  ratios,  and  the  fundamental  relations  of  the  functions  are  estab- 
lished and  illustrated  by  numerous  examples.     It  is   afterwards 


shown  how  the  numerical  values  of  the  ratios  may  be  represented 
by  lines,  and  the  simpler  line  values  are  employed  in  studying  the 
changes  of  the  functions  as  the  angle  changes. 

In  the  second  chapter  the  right  triangle  is  solved,  and  many 
problems  are  given  in  order  that  the  student  may  at  the  outset  per- 
ceive the  practical  utility  of  Trigonometry,  and  acquire  skill  in  the 
use  of  logarithms. 

In  the  third  chapter  the  definitions  of  the  functions  are  extended 
to  all  angles,  and  the  necessary  propositions  are  established  by  sim- 
ple proofs. 

In  the  fourth  and  last  chapter  the  oblique  triangle  is  solved,  and 
a  collection  of  miscellaneous  examples  is  added. 

The  answers  to  the  problems  are  printed  at  the  end  of  the  book. 


Wentworth's  Surueying. 

Paper.     Introduction,  25  cts. 

Wentworth's  Plane  and  Spherical  Trigonometry, 

Surveying,  and  Navigation. 

Introduction,  $1.12;  Allowance  for  old  book  in  use,  40  cts. 

The  object  of  this  brief  work  on  Surveying  and  Navigation  is  to 
present  these  subjects  in  a  clear  and  intelligible  way,  according 
to  the  best  methods  in  actual  use ;  and  also  to  present  them  in  so 
small  a  compass,  that  students  in  general  may  find  the  time  to 
acquire  a  competent  knowledge  of  these  very  interesting  and  im- 
portant studies. 

Wentworth's  Plane  and  Spherical  Trigonometry, 

and  Surveying.       With  Tables. 

Introduction,  $1.25;  Allowance  for  old  book  in  use,  40  cts. 


Wentworth  &  Hill's  Five-Place  Logarithmic  and 

Trigonometric  Tables.      (Seven  Tables.) 

Introduction,  50  cts. 

Wentworth  &  Hill's  Five-Place  Logarithmic  and 

Trigonometric  Tables.      (Complete  Edition.) 

Introduction,  $1.00. 

These  tables  have  been  prepared  mainly  from  Gauss's  Tables, 
and  are  designed  for  the  use  of  schools  and  colleges. 

Table  I.  contains  the  common  logarithms  of  the  natural  numbers 
from  1  to  10,000. 

Table  II.  contains  the  values  of  ir,  its  most  useful  combinations, 
and  the  corresponding  logarithms. 

Table  III.  contains  the  logarithms  of  the  trigonometric  functions 
of  angles  from  0°  to  0°  3'  and  from  89°  57'  to  90°  for  every  second ; 
from  0°  to  2°  and  from  88°  to  90°  for  every  10  seconds  ;  and  from 
1°  to  89°  for  every  minute. 

Table  IV.  gives  a  method  of  working  with  great  accuracy  when 
the  angle  lies  between  0°  and  2°  or  88°  and  90°. 

Table  V.  contains  the  natural  sines,  cosines,  tangents,  and  cotan- 
gents to  four  decimal  places,  and  at  intervals  of  10  minutes. 

Table  VI.  contains  the  values  of  the  circumference  and  area  of 
a  circle  for  different  values  of  the  radius,  and  of  the  radius  and 
area  for  different  values  of  the  circumference. 

The  tables  are  preceded  by  an  introduction,  in  which  the  nature 
and  use  of  logarithms  are  explained,  and  all  necessary  instruction 
given  for  using  the  tables. 

The  tables  occupy  60  pages  and  are  printed  in  large  type  with 
very  open  spacing.  Compactness,  simple  arrangement,  and  figures 
large  enough  not  to  strain  the  eyes  are  secured  by  excluding  pro- 
portional parts  from  the  tables.  These  are  considerations  of  the 
very  highest  importance,  and  it  is  doubtful  whether  the  printing  of 


proportional  parts  has  any  advantage  for  the  purposes  of  instruction 
where  the  main  object  is  to  inculcate  principles.  Experience  shows 
that  beginners  without  the  aid  of  proportional  parts  learn  in  a  very 
short  time  to  interpolate  with  great  rapidity  and  accuracy. 

Since  so  many  wish  these  Tables  separate,  we  have  published 
them  in  convenient  form,  at  a  price  hardly  covering  the  cost  of 
manufacture. 

Wentworth  &  Hill's  Examination  Manual. 

I.  Arithmetic.  Introduction,  35  cts. 

Wentworth  &  Hill's  Examination  Manual. 

II.  Algebra.  Introduction,  85  cts. 

Wentworth  &  Hill's  Exercise  Manual. 

II.   Algebra.  Introduction,  85  cts. 

(The  last  two  may  be  had  in  one  volume.) 


Wentworth  &  Hill's  Exercise  Manual  of  Arithmetic. 

In  Press. 

Wentworth  &  Hill's  Exercise  Manual  of  Geometry. 

In  Press. 

These,  and  others  to  follow,  are  a  series  of  short  Manuals,  intended 
to  cover  the  main  subjects  studied  in  our  schools  and  colleges. 
Each  Manual  is  confined  to  one  subject,  and  consists  of  two  parts : 
the  first  containing  about  100  examination  papers  made  from  the 
best  collections  of  questions ;  the  second  containing  recent  papers 
actually  set  in  English  and  American  schools  and  colleges.  Each 
Manual  also  contains  a  paper  completely  worked  out,  as  a  model. 


Mathematical  Books. 


INTROD.  PRICK. 

Byerly Differential  Calculus      $2.00 

Integral  Calculus 2.00 

Syllabus  of  Plane  Trigonometry 10 

Syllabus  of  Analytical  Geometry 10 

Syllabus  of  Analytical  Geometry,  adv.  course    .  10 
Syllabus  of  Equations 10 

G1°» Additio"TabIets{!argetsSe:  ".   ".  \   \  \    SM 

II  ulsted Mensuration 1.00 

Hardy Quaternions 2.00 

Hill Geometry  for  Beginners 1.00 

Peirce Three  and  Four-Place  Logarithms 40 

Tables,  chiefly  to  Four  Figures 40 

Elements  of  Logarithms 50 

Tables  of  Integrals 10 

Waldo Multiplication  and  Division  Tables  :  — 

Folio  size 50 

Small  size 25 

Wentworth  .    .    .  Elements  of  Algebra 1.12 

Complete  Algebra 1.40 

Plane  Geometry 75 

Plane  and  Solid  Geometry 1.25 

Plane  and  Solid  Geometry  and  Trigonometry  1.40 

Plane  Trigonometry.     Paper 30 

Plane  Trigonometry  and  Tables.    Paper    .      .60 

Plane  and  Spherical  Trigonometry 75 

Plane  and  Spherical  Trigonometry,  Survey- 
ing, and  Navigation  1.12 

Plane  and  Spherical  Trig,  and  Surveying, 

with  Tables 1.25 

Surveying.     Paper 25 

Trigonometric  Formulas 1.00 

Wentworth  &  Hill :  Five-Place  Log.  and  Trig.  Tables  (7  Tables)    .50 
Five-Place  Log.  and  Trig.  Tables  (  Comp.  Ed.)  1.00 

Practical  Arithmetic 1.00 

Examination  Manuals.    I.  Arithmetic     .    .      .35 
II.  Algebra      ...       .35 
Exercise  Manuals.     I.  Arithmetic    .... 

II.  Algebra 70 

III.  Geometry 

Wheeler Plane  and  Spherical  Trig,  and  Tables  •    .    .      1.00 

Copies  sent  to  Teachers  for  Examination,  with  a  view  to 
Introduction,  on  receipt  of  Introduction  Price, 


GINN,  HEATH,  &  CO.,  Publishers. 

BOSTON.  NEW  YORK.  CHICAGO. 


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